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# Electrostatics: Electric Charges and Fields

Sanjeev Dawar Classes

by

Tweet## Tarundeep Singh Bhatia

on 29 January 2013#### Transcript of Electrostatics: Electric Charges and Fields

Sanjeev Dawar Classes Electrostatics:

Electric Charges

& Fields Chapter 1 Physics Class 12 Contents of the Chapter 1. Introduction 2. Charge & its properties 3. Coulomb's Law 4. Coloumb's Law

(in Vector form) 5. Principle of

Superposition of charges 6. Electric Field 7. Principle of

Superposition of

Electric Field 8. Electric Dipole 9. Gauss's Theorem 10. Solved and

Unsolved Questions

and Numericals Introduction Charge and its Properties 1. Charge generally means a charged body. It can be positively ( + )

charged or negatively ( - ) charged.

2. If it is a positively charged body, then electrons are taken

out from the body and if negatively charged, electrons are added into

the body.

3. There are two types of charges positive charge (+q) and negative

charge (-q).

4. Like charges repel each other and unlike charges attract each

other.

q1.q2 > 0 (repulsion) q1.q2 < 0 (attraction)

5. Charge Develops on Body by

(a) Rubbing (friction)

(b) Conduction

(c) Induction

6. The S.I. unit of charge is Coloumb (C). It is a scalar quantity and is

additive in nature.

6. When the size of charged bodies is much smaller as compared

to distance between them they are termed as point charges.

7. Charges remain unaffected by motion that is

charge at rest = charges in motion

8. Conservation of Charge: Charge can neither be created nor be

destroyed but it can be transferred from

one part of the system to other part.

OR

The total charge of an isolated system must remain conserved.

9. Quantisation of charge: The total charge on a body is integral multiple

of fundamental charge ‘e’ ie.

q = ne

where n is an integer (n = 1, 2 , 3.....)

and Charge and Its Properties Coulomb's Law Coulomb's Law " The force of attraction or repulsion between two point charges is directly proportional to

the product of the magnitude of two charges and inversely proportional to the square of distance between

them. Also, the direction of the force is along the line joining the centres of the two charges." It states that: Mathematically, & ( Sign of q1 and q2 NOT to

be considered ) where, Coulomb's Law If charges are placed in an insulating medium of permittivity

then, where is absolute permittivity of medium

is relative permittivity of medium

unit less constant = 1 for free space, ( ≥ 1 always)

= 80 for water

= ∞ for metals Note: Electric Dipole is also known as DIELECTRIC CONSTANT.

The dielectric constant of a medium is defined as

" The ratio of permittivity of medium to the permittivity of free space" ,

that is, and so this implies : NOW, Coulomb's law (in vector form) One Coulomb charge..?? If two charges are in free space,

then, 1 Coulomb charge is the charge which when placed at a distance of 1 metre from an equal and similar charge in vacuum (or air) will repel it with a force of 9 x 10^9 N. Substituting the values, Let Coulomb's Law ( In vector form ): " Consider two like charges q1 and q2, located at ponits A and B in vacuum. The separation between the charges is 'r'. As charges are like, they repel each other. Let F21, be the force exerted on the charge q2 by charge q1 and F12, be the force that is exerted on charge q1 by charge q2. If r21 is the position vector of q2 relative to q1 and rˆ21 is unit vector along A to B, then the force is F21 is along A to B. " So, But, Similarly if rˆ12 is position vector of q1 relative to q2 and rˆ12 is unit vector from A to B. Obviously rˆ12 = - rˆ21, therefore equation (2) becomes ....(1) ....(2) ....(3) Comparing (1) and (3), we get This means that the Coulomb’s force exerted on q2 by q1 is equal and opposite to the Coulomb’s force exerted on q1 by q2, in accordance with Newtons third law.

Thus, Newtons third law also holds good for electrical forces This implies, 1. Branch of Physics which deals with the study of charges at rest is called

Electrostatics. 2. It deals with the study of forces, fields and potentials arising from static charges Principle of Superposition of Charges Coulomb’s law gives the force between two point charges. But if there are a number of interacting charges, then the force on a particular charge may be found by the principle of superpositionwhich states:

" If the system contains a number of interacting charges, then the force on a given charge is equal to the vector sum of the forces exerted on it by all remaining charges. "

It may be kept in mind that the force between any two charges is not affected by the presence of other charges. Principle of Superposition of Charges Suppose that a system of charges contains n charges q1, q2, q2,...qn having position vectors

relative to origin O respectively. A point charge q is located at P having position vector relative to O. The total force on q due to all n charges is to be found.

If F1, F2, F3,...... Fn are the forces acting on

q due to charges q1, q2, q3,...qn respectively, then by the principle of superposition, the net force on q is: If the force exerted due to charge qi on q is F, then from Coulomb’s law in vector form, The total force on q due to all the charges may be expressed as: Here represents the vector sum Electric Fields Electric Field " The space or region in which a charge experiences a force is called the electric field " To find the electric field strength

at a point in the electric field,

we place an infinitesimal

positive charge 'q0' at that point.

This charge is called the test charge.

The force acting on this test charge

is noted and this force divided by

the test charge gives electric field strength. The test charge is assumed so small that it does not cause any change in initial electric field.

Accordingly the electric field strength

is defined as follows:

" The electric field strength at any

point in an electric field is a vector quantity whose magnitude is

equal to the force acting an per

unit positive test charge and whose direction is along the direction of

force. " Electric Field If F is the force acting on positive test charge q0, then electric field strength is Here it has been assumed that test charge q0 is infinitesimal, therefore above definition may be expressed as The unit of electric field strength is Newton/Coulomb or Volt/Metre (abbreviated as N/C or V/m respectively). (i) The electric field strength due to a

point charge q at a distance r is: (ii) The electric field strength due to a system of discrete charge is : In vector form, Expression for Electric Field due to a point charge Consider a test charge q0 placed at a point P situated at a distance r from a point charge q. The force experienced by test charge Electric field intensity at given point Vectorially, That is Numericals on Coulomb's Law Question: A polythene piece rubbed with wool is found to have a negative charge of 3 x 10^-7 C.

(a) Estimate the number of electrons transferred. Also state from which to which the transfer of electron took place.

(b) Is there a transfer of mass from wool to polythene? Q1. What is the force between two small charged spheres having charges of 2 x 10^-7 C and 3 x 10^-7C

placed 30 cm apart in air ?

Q2. The electrostatic force on a small sphere of charge 0.4mC due to another small sphere of charge –0.8mC in air is 0.2N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?

Q3. (a) Two insulated charged copper spheres A and B have their centres separated by a distance of

50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 x 10^-7 C?

The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the

distance between them is halved?

Q4. Suppose the spheres A and B in the above have identical sizes. A third sphere of the same size

but uncharged is brought in contact with the first, then brought in contact with the second, and

finally removed from both. What is the new force of repulsion between A and B?

Q5. Two identical metallic spheres, having unequal opposite chargese are placed at a distance of

0.50m apart in air. After bringing them in contact with each other, they are again placed at the

same distance apart. Now the force of repulsion between them is 0.108 N. Calculate the final charge on each of them. Numericals on Coulomb's Law NUMERICALS

ON SUPERPOSITION PRINCIPLE OF

CHARGES Q1. A charge q is placed at the centre of the line joining two equal charges Q. Show that

the system of three charges charges will be in equilibrium of q= - Q/4.

Q2. Two similar charges, one twice in magnitude of other are placed 10cm apart.

Determine the position where a unit +ve charge should be placed on the line joining

the centres of two charges such that resultant force on unit +ve charge is zero.

Q3. Two fixed point charges +4e and e units are separated by a distance of a where

should a third point charge be place for it to be in equilibrium.

Q4. Two charges each of +q are placed along a line. A third charge Q is to be placed on

the line joining the two charges at what position and for value and sign of Q will the

system be in equilibrium.

Q5. Charges of +2, +2, –2μmicro coulomb are located at the vertices of equilateral triangle

of side 0.3m each find the net force experienced by each charge.

Q6. Three point charges of +2 mC, –3mC, –3mC are placed at the corners of the equilateral

triangle of side 20 cm each as shown what should be the magnitude and sign of the

charge to be placed at the mid point M of side BC so that the charge at A remains in

equilibrium.

Q7. Consider three charges q1, q2, q3 each equal to q placed at the vertices of an

equilateral triangle of side l, what is the force on a charge Q (with same sign as q)

placed at the centroid of the triangle.

Q8. Four point charge qA = 2 mC, qB = –5μmC, qC = 2μmC & qD = –5μmC are located at

the corners of a square ABCD of side 10 cm. What is the force on a charge of 1μC

placed at the centre of the sphere.

Q9. A charge Q, located at a point r, r is in equilibrium under the combined electric field

of three charges q1, q2, q3. If the charges q1, q2 are located at point r1 and r2

respectively, find the direction of the force on Q due to q3, in terms of q1, q2, r1, r2

and r . NOTE : 1. A charge is always placed in its own electric field but a charge never

experiences a force due to its own electric field. It experiences a

force only when it is placed in the electric field produced by some

other charge.

2. A Static charge or moving charge both are source of same electric

field.

3. Electric field is a vector quantity and direction of electric field is the

direction along which a small positive charge tends to move in and

electric field.

4. Dimensions:

5. Electric field due to a given charge depends on the space

coordinates r but the field exists at every point in 3D space which

means the electric field is continuous in nature. Moreover the

magnitude of electric field ‘E’ due to the point charge is same on a

sphere with the point charge at its centre i.e. electric field is

spherically symmetrical. Electric Field Lines Electric Field Lines " It is a line straight or curved drawn to represent an electric field such that tangent at any pt on it gives the direction of electric field at that point. " Properties of Electric Field Lines (1) A field line starts from the +ve charge and ends at the -ve charge.

(2) Tangent at a pt. on a field line gives the direction of electric field at

that point.

(3) Two field lines can never cross each other, if they did, then there

would be two tangent at the pt of intensection. Hence, two directions

of electric field at the same pt which is not possible.

(4) Field lines are drawn such that no. of field lines crossing per unit area

is directly proportional to the magnitude of electric field at that pt.

Hence, field lines come closer to each other where electric field is

higher and get away from each other where electric field is small. (5) Field lines are always perpendicular to the surface of the conductors.

(6) The field lines do not pass through the body of a conductor.

(7) It can not form closed loops.

(8) Electric field lines are continuous without any break.

(9) The electric lines of force have a tendency to contract lengthwise like

a stretched elastic string and separate from each other laterally. The

reason is that opposite charges attract and similar charges repel.

(10) The equidistant electric lines of force represent uniform electric field

while electric lines of forces at different separations represent non-

uniform electric field. Uniform Electric Field Non - Uniform Electric Field The above images display the electric field lines for some typical charges, conductors and dipoles. Question: Sketch the Electric Field Lines due to :

(a) Positive charge

(b) Negative Charge

(c) Electric Dipole

(d) Infinite Charge Sheet

(e) Uniform Electric Field

(f) Due to group of two point charges q1, q2

(i) q1 = q2

(ii) q1 > q2

(iii)q1 < q2

(g) Two uniformly positively charged plates parallel to each other

(h) Two oppositely charged plates parallel to each other

(i) Uniformly charged spherical shell

(j) Uncharged conducting sphere placed in uniform electric field

(k) Two equal negative charges placed at a certain distance apart. PRINCIPLE OF SUPER POSITION OF ELECTRIC FIELD NUMERICALS ON ELECTRIC FIELD "

The net electric field at a given point 'p' due to a number of charges (q1, q2, q3) is equal to the vector sum of all the electric fields produced by different charges.

" Principle of Superposition of Electric Fields that is, Numericals on Electric Field Q1. Two point charges of +5 x 10^-19 C and +20 x 10 ^-19 C are

separated by a distance of 2 m. Find the point on the line

joining them at which electric field intensity is zero.

Q2. A stream of electrons travelling with speed v m/s at right angles

to a uniform electric field E is deflected in a circular path of

radius r. Prove that

Q3. A particle of mass m and charge (–q) enters the region between

the two charged plates initially moving along X axis with speed

V as shown in fig. The length of each plate is L and an uniform

electric field e is mainitained between the plates. Show that the

vertical deflection of the particle at the far edge of the plate is

Q4. Suppose that the particle in above exercise is an electron

projected with velocity V = 2.0 x 10^6 m/s. If electric field

between the plates separated by 0.5 cm is 9.1 x 10^2 N/C, at

what maximum distance will the electron strike the above

plate? (e=1.6 x 10^–19C, mass of e = 9.1 x 10^–31 kg.)

Q5. An oil drop of 12 excess electrons is held stationary under a

constant electric field of 2.55 x 10^4 N/C. If the density of oil is

1.26g/cm. Estimate the size of the drop. Numericals on Electric Field Q6. Two point charges 3 mC and –3 mC are located

20cm apart.

(1) What is the electric field at the mid pt of the

line AB.(2) If a negative test charge of magnitude 1.5 x

10^-9C is placed at this pt. what is the force

experienced by the test charge.(3) Two point charges of

+5 x 10^-19 C and +20 x 10^-19C are separated by a

distance of 2 m. Find the point on the line joining them

at which electric field intensity is zero.

Q7. An electron falls through a distance of 1.5 cm in a

uniform electric field of magnitude 2.0 x 10^4 N/C. The

direction of the field is reversed keeping its magnitude

unchanged and a proton falls through the same

distance. Compute the time of fall in each case.

Contrast the situation with that of ‘free fall under

gravity’.

Q8. Two point charged q1 and q2, of magnitude +10^–8 C

and –10^–8C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C. "

When a charge q0 is placed in an external

electric field E, it experiences a force F = qE. If

q0 is positive then, force will be in the same

direction as that of electric field & if it is

negative then, F will be in opposite direction to

that of E.

" NOTE: A system containing two equal and opposite charges separated by a finite distance is called an electric dipole.

Dipole moment of electric dipole having charges +q and –q at separation 2l is. Electric Dipole 1. It is a vector quantity, directed from –q to +q.

Remark: Net charge on an electric dipole is zero 2. Its S.I. unit is cm Electric Field On The Axial Line Of Dipole ELECTRIC FIELD ON THE AXIAL LINE OF DIPOLE Let E1 and E2 be the electric field strengths at point P due to charges +q and –q respectively. Therefore (from B to P) (from P to A) Clearly the directions of electric field strengths E1 and E2 are along the same line but opposite to each other and E1 > E2 because positive charge is nearer.

The resultant electric field due to electric dipole has magnitude equal to the difference of E1 and E2 and direction from B to P i.e. But q.1l=p (electric dipole moment) ... ( 1 ) If the dipole is infinitely small and point P is far away from the dipole, the r >>l, therefore equation (1) may be expressed as OR ... ( 2 ) Electric Field On The Equatorial Point Consider a point P on the equatorial position of dipole formed of charges +q and –q at separation 2l. The distance of point P from mid point (O) of electric dipole is r. Let E1 and E2 be the electric field strengths due to charges +q and –q of electric dipole. From fig. along B to P along P to A Clearly, E1 and E2 are equal in magnitude i.e., Resolving the components of E1, E2 : E1 sin , E2 sin being equal and opposite cancel each other whereas E1 cos and E2 cos add up to give resultant electric field. Electric Field On The Equatorial Point Therefore,

Resultant electric field at P is But, and therefore, But q2l = p = electric dipole moment ... ( 3 ) If dipole is infinitestimal and point P is far away, we have l << r, so l^2 may be neglected as compared to r^2 and so equation (3) gives, i.e. electric field strength due to a short dipole at broadside on position

parallel to BA Please Note:

Its direction is parallel to the axis of dipole from positive to negative charge. (1) Electric field strength due to a short dipole at any

point is inversely proportional to the cube of its

distance from the dipole and the electric field strength

at axial position is twice that at broad-side on position

for the same distance. Note: (2) BEHAVIOUR OF AN ELECTRIC DIPOLE IN A UNIFORM ELECTRIC FIELD BEHAVIOUR OF AN ELECTRIC DIPOLE IN A UNIFORM ELECTRIC FIELD Consider an electric dipole placed in a uniform electric field of strength E in such a way that its dipole moment p makes an angle with the direction of E. The charges of dipole are –q and +q at separation 2l the dipole moment of electric dipole. p = q2l The force on charge +q is, F1 = qE, along the direction of field E. The force on charge –q is, F = qE, opposite to the direction of field E.

Obviously forces F1 and F2 are equal in magnitude but opposite in direction; hence net force on electric dipole in uniform electric field is: Force: As net force on electric dipole is zero, so dipole does not undergo any translatory

motion. Torque: The forces F1 and F2 form a couple (or torque) which tends to rotate and align the dipole along the direction of electric field. This couple is called the torque and is denoted by .

therefore, torque = magnitude of one force x

perpendicular distance between lines of

action of forces Clearly, the magnitude of torque depends on orientation ( θ) of the electric dipole relative to

electric field.

Torque ( τ) is a vector quantity whose direction is perpendicular to both p and E.

In vector form Thus, if an electric dipole is placed in an electric field in oblique orientation, it experiences no force but experiences a torque. The torque to align the dipole moment along the direction of electric field. Thus, if an electric dipole is placed in an electric field in oblique orientation, it experiences no force but experiences a torque. The torque to aligns the dipole moment along the direction of electric field.

Maximum Torque:

For maximum torque sin should be the maximum. As the

maximum value of sin = 1 when = 90º

Maximum torque,

Special Cases :

The torque experienced by the dipole is max. when held perpendicular to the electric field & is min. when it is

(i) Parallel ( = 0 )

(ii) Anti parallel ( = 180 )

to the direction of field. WORK DONE IN ROTATING THE DIPOLE IN AN ELECTRIC FIELD

(POTENTIAL ENERGY OF A DIPOLE) Let an electric dipole be rotated in electric field from angle to in the direction of electric field. In this process the angle of orientation is changing continuously; hence the torque also changes continuously.

Let at any time, the angle between dipole moment p and electric field E be , then torque on dipole

The work done in rotating the dipole a further small angle d is

dW = Torque x angular displacement

Total work done in rotating the dipole from

angle and will be the sum of all such

small works which is found by integration WORK DONE IN ROTATING THE DIPOLE IN AN ELECTRIC FIELD

(POTENTIAL ENERGY OF A DIPOLE) Special Case: (1) If electric dipole is initially in a stable equilibrium

position ( = 0 ) and rotated through an angle ( = ),

then work done (2) Work done in rotating a dipole from end to end. Stable equilibrium: Unstable equilibrium: Few note worthy points: 1. In a non uniform electric field the

electric dipole experiences a net

force as well as a net torque.

2. Electric field due to a dipole

doesn’t exhibit spherical

symmetry because electric field at

a point on a axial line is double as

compared to that a corresponding

point along the equatorial line

infact electric field due to a dipole

exhibits cyclindrical symmetry. Distribution of Charge

(Continuous Charge Distribution) Distribution of Charge:

(Continuous Charge distribution) A continuous charge distribution means a

system of extra large number of closely

spaced charges & it may be of following

types: 1. Line Charge Distribution:

In which charge is distributed uniformly

along a line.

eg. a charged wire, ring etc.

If represents the linear charge density

around a length element represented by r,

then,

2. Surface Distribution of Charge:

It is a continuous distribution of charge over

some area.

eg. a thin charged sheet.

If represents the surface charge density

along a surface element, then, 3. Volume Charge distribution:

The continuous charge distribution

along a given volume.

If represents the volume charge density around a volume ‘dv’ then, Electric Flux Electric Flux: The number of electric lines of force diverging normally from a surface is called the electric flux through that surface. Electric flux through surface element is

, where E is the electric field strength 1. Electric Flux is said to be positive, if lines are

following outwards i.e. < 90º

2. Electric Flux is said to be negative, if lines are

following inwards i.e. > 90º

3. Electric Flux is said to be zero when surface is

held normal to the direction of E i.e. = 90º

4. Electric flux is a scalar quantity because it is a

scalar product of E & ds.

5. Significance:

It gives the total number of electric field lines

that cross a given surface normally. Question: Q1. For what position of an electric dipole is the electric field

(i) Parallel to the line joining the two charges (equatorial position)

(ii) Along the line joining the two charges

(axial position). Some More Questions: Q2. Two charges ± 10μmC are placed 5.0 mm apart.

Determine the electric field at

(a) a point P on the axis of the dipole 15 cm

away from its centre O on the side of the

positive charge, as shown in fig.(a), and

(b) a point 15 cm away from O on a line

passing through O and normal to the axis

of the dipole, as shown in fig.(b). Q3. The following data was obtained for the

dependence of the magnitude of the electric

field,with distance, from a reference point, O,

within the charge distribution in the shaded

region.

(i) Identify the charge distribution and justify your

answer.

(ii) If the potential due to this charge distribution,

has a value V at the point A, what is its value

at the point A’? Question: Q4. What is the angle between the directions of electric field at any (i) axial point and

(ii) equatorial point due to an electric dipole? Question: Q. An electric dipole of dipole moment p is placed in a uniform electric field E. Write the expression for the torque experienced by the dipole. Identify two pairs of perpendicular vectors in the expression.

Show diagramatically the orientation of the dipole in the field for which the torque is

(i) maximum

(ii) half the maximum value

(iii) zero Questions Questions: Q1. An electric dipole consist of two opposite

charges each of 1 micro coulomb, separated

by 2 cm. Dipole is placed in an external

uniform field of 10^5 N/C. Find

(i) Torque, when dipole makes an angle 60

degrees with the electric field.

(ii) Max. torque.

(iii) Work done in rotating the dipole through 30

degrees starting from position = 0

(iv)Work done in rotating the dipole through 180

degrees starting from the position = 0 Q2. An electric dipole of length 10 cm having charges

6 x 10^–3 C, placed at 30 degrees with respect to a

uniform electric field, experience a torque of magnitude

Nm. Calculate

(i) magnitude of electric field and

(ii) potential energy of electric dipole.

Q3. A system has two charges qA = 2.5 x 10^-7 C and

qB= – 2.5 x 10^-7 C located at point A = (0, 0, –15 cm) and

electric dipole moment of the system. Electric flux through entire closed surface is S.I. unit of electric flux is volt metre or Nm2/C

Full transcriptElectric Charges

& Fields Chapter 1 Physics Class 12 Contents of the Chapter 1. Introduction 2. Charge & its properties 3. Coulomb's Law 4. Coloumb's Law

(in Vector form) 5. Principle of

Superposition of charges 6. Electric Field 7. Principle of

Superposition of

Electric Field 8. Electric Dipole 9. Gauss's Theorem 10. Solved and

Unsolved Questions

and Numericals Introduction Charge and its Properties 1. Charge generally means a charged body. It can be positively ( + )

charged or negatively ( - ) charged.

2. If it is a positively charged body, then electrons are taken

out from the body and if negatively charged, electrons are added into

the body.

3. There are two types of charges positive charge (+q) and negative

charge (-q).

4. Like charges repel each other and unlike charges attract each

other.

q1.q2 > 0 (repulsion) q1.q2 < 0 (attraction)

5. Charge Develops on Body by

(a) Rubbing (friction)

(b) Conduction

(c) Induction

6. The S.I. unit of charge is Coloumb (C). It is a scalar quantity and is

additive in nature.

6. When the size of charged bodies is much smaller as compared

to distance between them they are termed as point charges.

7. Charges remain unaffected by motion that is

charge at rest = charges in motion

8. Conservation of Charge: Charge can neither be created nor be

destroyed but it can be transferred from

one part of the system to other part.

OR

The total charge of an isolated system must remain conserved.

9. Quantisation of charge: The total charge on a body is integral multiple

of fundamental charge ‘e’ ie.

q = ne

where n is an integer (n = 1, 2 , 3.....)

and Charge and Its Properties Coulomb's Law Coulomb's Law " The force of attraction or repulsion between two point charges is directly proportional to

the product of the magnitude of two charges and inversely proportional to the square of distance between

them. Also, the direction of the force is along the line joining the centres of the two charges." It states that: Mathematically, & ( Sign of q1 and q2 NOT to

be considered ) where, Coulomb's Law If charges are placed in an insulating medium of permittivity

then, where is absolute permittivity of medium

is relative permittivity of medium

unit less constant = 1 for free space, ( ≥ 1 always)

= 80 for water

= ∞ for metals Note: Electric Dipole is also known as DIELECTRIC CONSTANT.

The dielectric constant of a medium is defined as

" The ratio of permittivity of medium to the permittivity of free space" ,

that is, and so this implies : NOW, Coulomb's law (in vector form) One Coulomb charge..?? If two charges are in free space,

then, 1 Coulomb charge is the charge which when placed at a distance of 1 metre from an equal and similar charge in vacuum (or air) will repel it with a force of 9 x 10^9 N. Substituting the values, Let Coulomb's Law ( In vector form ): " Consider two like charges q1 and q2, located at ponits A and B in vacuum. The separation between the charges is 'r'. As charges are like, they repel each other. Let F21, be the force exerted on the charge q2 by charge q1 and F12, be the force that is exerted on charge q1 by charge q2. If r21 is the position vector of q2 relative to q1 and rˆ21 is unit vector along A to B, then the force is F21 is along A to B. " So, But, Similarly if rˆ12 is position vector of q1 relative to q2 and rˆ12 is unit vector from A to B. Obviously rˆ12 = - rˆ21, therefore equation (2) becomes ....(1) ....(2) ....(3) Comparing (1) and (3), we get This means that the Coulomb’s force exerted on q2 by q1 is equal and opposite to the Coulomb’s force exerted on q1 by q2, in accordance with Newtons third law.

Thus, Newtons third law also holds good for electrical forces This implies, 1. Branch of Physics which deals with the study of charges at rest is called

Electrostatics. 2. It deals with the study of forces, fields and potentials arising from static charges Principle of Superposition of Charges Coulomb’s law gives the force between two point charges. But if there are a number of interacting charges, then the force on a particular charge may be found by the principle of superpositionwhich states:

" If the system contains a number of interacting charges, then the force on a given charge is equal to the vector sum of the forces exerted on it by all remaining charges. "

It may be kept in mind that the force between any two charges is not affected by the presence of other charges. Principle of Superposition of Charges Suppose that a system of charges contains n charges q1, q2, q2,...qn having position vectors

relative to origin O respectively. A point charge q is located at P having position vector relative to O. The total force on q due to all n charges is to be found.

If F1, F2, F3,...... Fn are the forces acting on

q due to charges q1, q2, q3,...qn respectively, then by the principle of superposition, the net force on q is: If the force exerted due to charge qi on q is F, then from Coulomb’s law in vector form, The total force on q due to all the charges may be expressed as: Here represents the vector sum Electric Fields Electric Field " The space or region in which a charge experiences a force is called the electric field " To find the electric field strength

at a point in the electric field,

we place an infinitesimal

positive charge 'q0' at that point.

This charge is called the test charge.

The force acting on this test charge

is noted and this force divided by

the test charge gives electric field strength. The test charge is assumed so small that it does not cause any change in initial electric field.

Accordingly the electric field strength

is defined as follows:

" The electric field strength at any

point in an electric field is a vector quantity whose magnitude is

equal to the force acting an per

unit positive test charge and whose direction is along the direction of

force. " Electric Field If F is the force acting on positive test charge q0, then electric field strength is Here it has been assumed that test charge q0 is infinitesimal, therefore above definition may be expressed as The unit of electric field strength is Newton/Coulomb or Volt/Metre (abbreviated as N/C or V/m respectively). (i) The electric field strength due to a

point charge q at a distance r is: (ii) The electric field strength due to a system of discrete charge is : In vector form, Expression for Electric Field due to a point charge Consider a test charge q0 placed at a point P situated at a distance r from a point charge q. The force experienced by test charge Electric field intensity at given point Vectorially, That is Numericals on Coulomb's Law Question: A polythene piece rubbed with wool is found to have a negative charge of 3 x 10^-7 C.

(a) Estimate the number of electrons transferred. Also state from which to which the transfer of electron took place.

(b) Is there a transfer of mass from wool to polythene? Q1. What is the force between two small charged spheres having charges of 2 x 10^-7 C and 3 x 10^-7C

placed 30 cm apart in air ?

Q2. The electrostatic force on a small sphere of charge 0.4mC due to another small sphere of charge –0.8mC in air is 0.2N.

(a) What is the distance between the two spheres?

(b) What is the force on the second sphere due to the first?

Q3. (a) Two insulated charged copper spheres A and B have their centres separated by a distance of

50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 x 10^-7 C?

The radii of A and B are negligible compared to the distance of separation.

(b) What is the force of repulsion if each sphere is charged double the above amount, and the

distance between them is halved?

Q4. Suppose the spheres A and B in the above have identical sizes. A third sphere of the same size

but uncharged is brought in contact with the first, then brought in contact with the second, and

finally removed from both. What is the new force of repulsion between A and B?

Q5. Two identical metallic spheres, having unequal opposite chargese are placed at a distance of

0.50m apart in air. After bringing them in contact with each other, they are again placed at the

same distance apart. Now the force of repulsion between them is 0.108 N. Calculate the final charge on each of them. Numericals on Coulomb's Law NUMERICALS

ON SUPERPOSITION PRINCIPLE OF

CHARGES Q1. A charge q is placed at the centre of the line joining two equal charges Q. Show that

the system of three charges charges will be in equilibrium of q= - Q/4.

Q2. Two similar charges, one twice in magnitude of other are placed 10cm apart.

Determine the position where a unit +ve charge should be placed on the line joining

the centres of two charges such that resultant force on unit +ve charge is zero.

Q3. Two fixed point charges +4e and e units are separated by a distance of a where

should a third point charge be place for it to be in equilibrium.

Q4. Two charges each of +q are placed along a line. A third charge Q is to be placed on

the line joining the two charges at what position and for value and sign of Q will the

system be in equilibrium.

Q5. Charges of +2, +2, –2μmicro coulomb are located at the vertices of equilateral triangle

of side 0.3m each find the net force experienced by each charge.

Q6. Three point charges of +2 mC, –3mC, –3mC are placed at the corners of the equilateral

triangle of side 20 cm each as shown what should be the magnitude and sign of the

charge to be placed at the mid point M of side BC so that the charge at A remains in

equilibrium.

Q7. Consider three charges q1, q2, q3 each equal to q placed at the vertices of an

equilateral triangle of side l, what is the force on a charge Q (with same sign as q)

placed at the centroid of the triangle.

Q8. Four point charge qA = 2 mC, qB = –5μmC, qC = 2μmC & qD = –5μmC are located at

the corners of a square ABCD of side 10 cm. What is the force on a charge of 1μC

placed at the centre of the sphere.

Q9. A charge Q, located at a point r, r is in equilibrium under the combined electric field

of three charges q1, q2, q3. If the charges q1, q2 are located at point r1 and r2

respectively, find the direction of the force on Q due to q3, in terms of q1, q2, r1, r2

and r . NOTE : 1. A charge is always placed in its own electric field but a charge never

experiences a force due to its own electric field. It experiences a

force only when it is placed in the electric field produced by some

other charge.

2. A Static charge or moving charge both are source of same electric

field.

3. Electric field is a vector quantity and direction of electric field is the

direction along which a small positive charge tends to move in and

electric field.

4. Dimensions:

5. Electric field due to a given charge depends on the space

coordinates r but the field exists at every point in 3D space which

means the electric field is continuous in nature. Moreover the

magnitude of electric field ‘E’ due to the point charge is same on a

sphere with the point charge at its centre i.e. electric field is

spherically symmetrical. Electric Field Lines Electric Field Lines " It is a line straight or curved drawn to represent an electric field such that tangent at any pt on it gives the direction of electric field at that point. " Properties of Electric Field Lines (1) A field line starts from the +ve charge and ends at the -ve charge.

(2) Tangent at a pt. on a field line gives the direction of electric field at

that point.

(3) Two field lines can never cross each other, if they did, then there

would be two tangent at the pt of intensection. Hence, two directions

of electric field at the same pt which is not possible.

(4) Field lines are drawn such that no. of field lines crossing per unit area

is directly proportional to the magnitude of electric field at that pt.

Hence, field lines come closer to each other where electric field is

higher and get away from each other where electric field is small. (5) Field lines are always perpendicular to the surface of the conductors.

(6) The field lines do not pass through the body of a conductor.

(7) It can not form closed loops.

(8) Electric field lines are continuous without any break.

(9) The electric lines of force have a tendency to contract lengthwise like

a stretched elastic string and separate from each other laterally. The

reason is that opposite charges attract and similar charges repel.

(10) The equidistant electric lines of force represent uniform electric field

while electric lines of forces at different separations represent non-

uniform electric field. Uniform Electric Field Non - Uniform Electric Field The above images display the electric field lines for some typical charges, conductors and dipoles. Question: Sketch the Electric Field Lines due to :

(a) Positive charge

(b) Negative Charge

(c) Electric Dipole

(d) Infinite Charge Sheet

(e) Uniform Electric Field

(f) Due to group of two point charges q1, q2

(i) q1 = q2

(ii) q1 > q2

(iii)q1 < q2

(g) Two uniformly positively charged plates parallel to each other

(h) Two oppositely charged plates parallel to each other

(i) Uniformly charged spherical shell

(j) Uncharged conducting sphere placed in uniform electric field

(k) Two equal negative charges placed at a certain distance apart. PRINCIPLE OF SUPER POSITION OF ELECTRIC FIELD NUMERICALS ON ELECTRIC FIELD "

The net electric field at a given point 'p' due to a number of charges (q1, q2, q3) is equal to the vector sum of all the electric fields produced by different charges.

" Principle of Superposition of Electric Fields that is, Numericals on Electric Field Q1. Two point charges of +5 x 10^-19 C and +20 x 10 ^-19 C are

separated by a distance of 2 m. Find the point on the line

joining them at which electric field intensity is zero.

Q2. A stream of electrons travelling with speed v m/s at right angles

to a uniform electric field E is deflected in a circular path of

radius r. Prove that

Q3. A particle of mass m and charge (–q) enters the region between

the two charged plates initially moving along X axis with speed

V as shown in fig. The length of each plate is L and an uniform

electric field e is mainitained between the plates. Show that the

vertical deflection of the particle at the far edge of the plate is

Q4. Suppose that the particle in above exercise is an electron

projected with velocity V = 2.0 x 10^6 m/s. If electric field

between the plates separated by 0.5 cm is 9.1 x 10^2 N/C, at

what maximum distance will the electron strike the above

plate? (e=1.6 x 10^–19C, mass of e = 9.1 x 10^–31 kg.)

Q5. An oil drop of 12 excess electrons is held stationary under a

constant electric field of 2.55 x 10^4 N/C. If the density of oil is

1.26g/cm. Estimate the size of the drop. Numericals on Electric Field Q6. Two point charges 3 mC and –3 mC are located

20cm apart.

(1) What is the electric field at the mid pt of the

line AB.(2) If a negative test charge of magnitude 1.5 x

10^-9C is placed at this pt. what is the force

experienced by the test charge.(3) Two point charges of

+5 x 10^-19 C and +20 x 10^-19C are separated by a

distance of 2 m. Find the point on the line joining them

at which electric field intensity is zero.

Q7. An electron falls through a distance of 1.5 cm in a

uniform electric field of magnitude 2.0 x 10^4 N/C. The

direction of the field is reversed keeping its magnitude

unchanged and a proton falls through the same

distance. Compute the time of fall in each case.

Contrast the situation with that of ‘free fall under

gravity’.

Q8. Two point charged q1 and q2, of magnitude +10^–8 C

and –10^–8C, respectively, are placed 0.1 m apart. Calculate the electric fields at points A, B and C. "

When a charge q0 is placed in an external

electric field E, it experiences a force F = qE. If

q0 is positive then, force will be in the same

direction as that of electric field & if it is

negative then, F will be in opposite direction to

that of E.

" NOTE: A system containing two equal and opposite charges separated by a finite distance is called an electric dipole.

Dipole moment of electric dipole having charges +q and –q at separation 2l is. Electric Dipole 1. It is a vector quantity, directed from –q to +q.

Remark: Net charge on an electric dipole is zero 2. Its S.I. unit is cm Electric Field On The Axial Line Of Dipole ELECTRIC FIELD ON THE AXIAL LINE OF DIPOLE Let E1 and E2 be the electric field strengths at point P due to charges +q and –q respectively. Therefore (from B to P) (from P to A) Clearly the directions of electric field strengths E1 and E2 are along the same line but opposite to each other and E1 > E2 because positive charge is nearer.

The resultant electric field due to electric dipole has magnitude equal to the difference of E1 and E2 and direction from B to P i.e. But q.1l=p (electric dipole moment) ... ( 1 ) If the dipole is infinitely small and point P is far away from the dipole, the r >>l, therefore equation (1) may be expressed as OR ... ( 2 ) Electric Field On The Equatorial Point Consider a point P on the equatorial position of dipole formed of charges +q and –q at separation 2l. The distance of point P from mid point (O) of electric dipole is r. Let E1 and E2 be the electric field strengths due to charges +q and –q of electric dipole. From fig. along B to P along P to A Clearly, E1 and E2 are equal in magnitude i.e., Resolving the components of E1, E2 : E1 sin , E2 sin being equal and opposite cancel each other whereas E1 cos and E2 cos add up to give resultant electric field. Electric Field On The Equatorial Point Therefore,

Resultant electric field at P is But, and therefore, But q2l = p = electric dipole moment ... ( 3 ) If dipole is infinitestimal and point P is far away, we have l << r, so l^2 may be neglected as compared to r^2 and so equation (3) gives, i.e. electric field strength due to a short dipole at broadside on position

parallel to BA Please Note:

Its direction is parallel to the axis of dipole from positive to negative charge. (1) Electric field strength due to a short dipole at any

point is inversely proportional to the cube of its

distance from the dipole and the electric field strength

at axial position is twice that at broad-side on position

for the same distance. Note: (2) BEHAVIOUR OF AN ELECTRIC DIPOLE IN A UNIFORM ELECTRIC FIELD BEHAVIOUR OF AN ELECTRIC DIPOLE IN A UNIFORM ELECTRIC FIELD Consider an electric dipole placed in a uniform electric field of strength E in such a way that its dipole moment p makes an angle with the direction of E. The charges of dipole are –q and +q at separation 2l the dipole moment of electric dipole. p = q2l The force on charge +q is, F1 = qE, along the direction of field E. The force on charge –q is, F = qE, opposite to the direction of field E.

Obviously forces F1 and F2 are equal in magnitude but opposite in direction; hence net force on electric dipole in uniform electric field is: Force: As net force on electric dipole is zero, so dipole does not undergo any translatory

motion. Torque: The forces F1 and F2 form a couple (or torque) which tends to rotate and align the dipole along the direction of electric field. This couple is called the torque and is denoted by .

therefore, torque = magnitude of one force x

perpendicular distance between lines of

action of forces Clearly, the magnitude of torque depends on orientation ( θ) of the electric dipole relative to

electric field.

Torque ( τ) is a vector quantity whose direction is perpendicular to both p and E.

In vector form Thus, if an electric dipole is placed in an electric field in oblique orientation, it experiences no force but experiences a torque. The torque to align the dipole moment along the direction of electric field. Thus, if an electric dipole is placed in an electric field in oblique orientation, it experiences no force but experiences a torque. The torque to aligns the dipole moment along the direction of electric field.

Maximum Torque:

For maximum torque sin should be the maximum. As the

maximum value of sin = 1 when = 90º

Maximum torque,

Special Cases :

The torque experienced by the dipole is max. when held perpendicular to the electric field & is min. when it is

(i) Parallel ( = 0 )

(ii) Anti parallel ( = 180 )

to the direction of field. WORK DONE IN ROTATING THE DIPOLE IN AN ELECTRIC FIELD

(POTENTIAL ENERGY OF A DIPOLE) Let an electric dipole be rotated in electric field from angle to in the direction of electric field. In this process the angle of orientation is changing continuously; hence the torque also changes continuously.

Let at any time, the angle between dipole moment p and electric field E be , then torque on dipole

The work done in rotating the dipole a further small angle d is

dW = Torque x angular displacement

Total work done in rotating the dipole from

angle and will be the sum of all such

small works which is found by integration WORK DONE IN ROTATING THE DIPOLE IN AN ELECTRIC FIELD

(POTENTIAL ENERGY OF A DIPOLE) Special Case: (1) If electric dipole is initially in a stable equilibrium

position ( = 0 ) and rotated through an angle ( = ),

then work done (2) Work done in rotating a dipole from end to end. Stable equilibrium: Unstable equilibrium: Few note worthy points: 1. In a non uniform electric field the

electric dipole experiences a net

force as well as a net torque.

2. Electric field due to a dipole

doesn’t exhibit spherical

symmetry because electric field at

a point on a axial line is double as

compared to that a corresponding

point along the equatorial line

infact electric field due to a dipole

exhibits cyclindrical symmetry. Distribution of Charge

(Continuous Charge Distribution) Distribution of Charge:

(Continuous Charge distribution) A continuous charge distribution means a

system of extra large number of closely

spaced charges & it may be of following

types: 1. Line Charge Distribution:

In which charge is distributed uniformly

along a line.

eg. a charged wire, ring etc.

If represents the linear charge density

around a length element represented by r,

then,

2. Surface Distribution of Charge:

It is a continuous distribution of charge over

some area.

eg. a thin charged sheet.

If represents the surface charge density

along a surface element, then, 3. Volume Charge distribution:

The continuous charge distribution

along a given volume.

If represents the volume charge density around a volume ‘dv’ then, Electric Flux Electric Flux: The number of electric lines of force diverging normally from a surface is called the electric flux through that surface. Electric flux through surface element is

, where E is the electric field strength 1. Electric Flux is said to be positive, if lines are

following outwards i.e. < 90º

2. Electric Flux is said to be negative, if lines are

following inwards i.e. > 90º

3. Electric Flux is said to be zero when surface is

held normal to the direction of E i.e. = 90º

4. Electric flux is a scalar quantity because it is a

scalar product of E & ds.

5. Significance:

It gives the total number of electric field lines

that cross a given surface normally. Question: Q1. For what position of an electric dipole is the electric field

(i) Parallel to the line joining the two charges (equatorial position)

(ii) Along the line joining the two charges

(axial position). Some More Questions: Q2. Two charges ± 10μmC are placed 5.0 mm apart.

Determine the electric field at

(a) a point P on the axis of the dipole 15 cm

away from its centre O on the side of the

positive charge, as shown in fig.(a), and

(b) a point 15 cm away from O on a line

passing through O and normal to the axis

of the dipole, as shown in fig.(b). Q3. The following data was obtained for the

dependence of the magnitude of the electric

field,with distance, from a reference point, O,

within the charge distribution in the shaded

region.

(i) Identify the charge distribution and justify your

answer.

(ii) If the potential due to this charge distribution,

has a value V at the point A, what is its value

at the point A’? Question: Q4. What is the angle between the directions of electric field at any (i) axial point and

(ii) equatorial point due to an electric dipole? Question: Q. An electric dipole of dipole moment p is placed in a uniform electric field E. Write the expression for the torque experienced by the dipole. Identify two pairs of perpendicular vectors in the expression.

Show diagramatically the orientation of the dipole in the field for which the torque is

(i) maximum

(ii) half the maximum value

(iii) zero Questions Questions: Q1. An electric dipole consist of two opposite

charges each of 1 micro coulomb, separated

by 2 cm. Dipole is placed in an external

uniform field of 10^5 N/C. Find

(i) Torque, when dipole makes an angle 60

degrees with the electric field.

(ii) Max. torque.

(iii) Work done in rotating the dipole through 30

degrees starting from position = 0

(iv)Work done in rotating the dipole through 180

degrees starting from the position = 0 Q2. An electric dipole of length 10 cm having charges

6 x 10^–3 C, placed at 30 degrees with respect to a

uniform electric field, experience a torque of magnitude

Nm. Calculate

(i) magnitude of electric field and

(ii) potential energy of electric dipole.

Q3. A system has two charges qA = 2.5 x 10^-7 C and

qB= – 2.5 x 10^-7 C located at point A = (0, 0, –15 cm) and

electric dipole moment of the system. Electric flux through entire closed surface is S.I. unit of electric flux is volt metre or Nm2/C