UNIT – 1 ELECTROSTATICS

ONE MARK QUESTIONS

1. A glass rod when rubbed with silk acquires a charge +1.6 × 10–12C. What about silk ?2. If Coulomb’s law involved 1/r3 dependence instead of 1/r2 , will the Gauss theorem be applicable?3. Define electric potential. Is it a vector or a scalar quantity? 4. Which orientation of an electric dipole in a uniform electric field would correspond to stable

equilibrium? 5. If the radius of the Gaussian surface enclosing a charge is halved, how does the electric flux through the

Gaussian surface change? 6. Define the electric dipole moment of a dipole. Write its SI unit. 7. What is the electrostatic potential due to an electric dipole at an equatorial point? 8. What is the work done in moving a test charge q through a distance of 1 cm along the equatorial axis

of an electric dipole 9. Define the term ‘potential energy’ of charge ‘q ‘at a distance ‘r’ in an external electric field.10. Name the physical quantity whose SI unit is J/C. is it scalar or vector quantity? 11. A hollow metal sphere of radius 5 cm is charged such that the potential on its surface is 10 V. what is

potential at the centre of the sphere? 12. A charge ‘q’ is placed at the centre of a cube of side l, what is the electric passing through each face of

the cube? 13. What is the electric flux passing through two opposite faces of the cube? 14. Two charges of magnitude -2Q and +Q are located at points (a,o) and (4a, o) respectively. What is the

electric flux due these charges through a sphere of radius ‘3a’ with its centre at the origin? 15. Two charges of magnitude -3Q and +2Q are located at points (a,o) and (4a, o) respectively. What is the

electric flux due these charges through a sphere of radius ‘5a’ with its centre at the origin? 16. Two equal balls having equal positive charge ’q’ coulombs are suspended by two strings of equal

length. What would be the effect on the force when a plastic sheet is inserted between the two? 17. A point charge +Q is placed in the vicinity of a conducting surface. Trace the field lines between the

charge and the conducting surface. 18. Name the physical quantities whose S.I. units are (i) coulomb/volt (ii) N/C (iii) V/m.

19. What would be the work done if a point charge +q is taken from a point at the circumference of a circle to another point at the circumference if another point charge is at the centre of the circle.

20. A and B are two conducting sphere of the same radii and same material. A being solid and another hollow. Both are charged to same potential. What will be relation between the capacitances of them ? which will assume more charge?

ANSWERS OF ONE MARK QUETIONS

Ans 1. Silk cloth with acquire equal and opposite charge i.e. –1.6 × 10–12 C

Ans 2. No, Gauss’law can be obtained from Coulomb’s law only if it has the 1/r2

Ans3. Electric potential at a point is equal to amount of work required to bring a unit positive charge from infinity to that point in the electric field. It is a scalar quantity.

Ans4 .When it is placed parallel to the electric field.

Ans5. Remain unchanged.

Ans6. It is defined as the product of either charge and the length of the dipole.

Mathematically, P= q (2a). Its SI unit is Cm.

Ans7. Zero.

Ans8. Zero.

Ans9. It is defined as the amount of work done in bringing the charge q from infinity to the location ‘r’ in an external electric field.

Ans10. Electrostatic potential. Scalar quantity

Ans11. 10 V

Ans12. Flux linked through each face of the cube =q/6ɛ0

Ans 13. Flux linked through two opposite faces of the cube=q/3 ɛ0

Ans14. Flux= -2Q/ ɛ0

Ans15. Flux= -Q/ ɛ0

Ans16. The will be reduced as force is inversely proportional to the dielectric constant K. for plastic K˃ 1.

Ans17.

Ans18. (i) Capacitance (ii) Electric field intensity (iii) Electric field intensity.

Ans19. Zero, because direction of motion is perpendicular to the force acting at the charge.

Ans20. Both will have same charges because capacitances in these cases are same.

TWO MARKS QUESTIONS

1. Four point charges 1μC, 1μC, -1μC and -1 μC are placed at corners of a square of each side 0.1 m (i) calculate electric potential at centre O of square (ii) if E is midpoint of BC, what is work done in carrying an electron from O to E? A B

E D C

2.Write an expression for potential energy of two charges q1 and q2 at r1 and r2 in a uniform electric field E.

3. A point charge Q is placed at the point O in following figure. Is the potential difference VA – VB positive, negative or zero if Q is (i) positive (ii) negative? Q----------------- A----------B

4. During lightning, the safest way to protect is to be inside the house or car. Why?5. Two capacitors of capacitances 6μF and 12μF are connected in series with a battery. The voltage

across 6μF capacitor is 2 V. compute the total supply voltage.6. Two charged spherical conductors of radii R1 and R2 when connected by a connecting wire acquire

charges q1 and q2 respectively. Find the ratio of their charge densities in the terms of their radii?7. Two point charges 4Q and Q are separated by 1 m in air. At what point on the line joining the

charges is the electric field intensity zero?Also calculate the electrostatic potential energy the system of charges, taking the value of Q= -2x10-7 C.

8. Draw the equipotential surfaces corresponding to a field that uniformly increases in magnitude but remain constant along Z-direction. How are these surfaces different from that of a constant electric field along Z-direction?

9. Define electric flux. Write its SI unit.A charge q is enclosed by a spherical surface of radius R. If the radius is reduced to half, how would the electric flux through the surface change?

10. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge ‘Q’. A charge ‘q’ is placed at the centre of the shell.a) What is the surface charge density on the (i) inner surface (ii) outer surface of the shell?b) Write the expression for the electric field at a point x˃r2 from the centre of the shell.

11. A thin straight infinitely long conducting wire having charge density λ is enclosed by a cylindrical surface of radius r and length l, its coinciding with the length of the wire. Find the expression for the electric flux through the surface of the cylinder.

12. Net capacitance of three identical capacitors in series is 3 μF. What is their net capacitance if connected in parallel? Find the ratio of the energy stored in the two configurations if they are both connected to the same source.

O

13. Plot a graph showing the variation of Coulomb force(F) versus (1/r2) where r is the distance between the two charges of each pair of charges (1 μC, 2μC ¿ and (2 μC, -3μC ¿ . interpret the graphs obtained.

14. A test charge ‘q’ is moved without acceleration from A to C along the path from A to B and then from B to C in electric field as shown in figure. (i) calculate the potential difference between A and C. (ii) at which point ( of the two) is the electric potential more and why?

15. An electric dipole is held in a uniform electric field. (i) Show that the net force acting on it is zero. (ii) The dipole is aligned parallel to the field. Find the work done in rotating it through the angle

of 1800. 16. A slab of material of dielectric constant k has the same area as that of the plates of a parallel plate

capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor.

17. A parallel plate capacitor of capacitance c is charged to a potential V. If it is then connected to another uncharged capacitor having the same capacitance. Find out ratio of the energy stored in the combined system to that stored initially in the single capacitor.

18. Derive an expression for the work done in rotating a dipole from the angle ɵo to ɵ1 on a uniform electric field E.

ANSWERS OF TWO MARKS QUESTIONS

Ans1- (0,0)

Ans 2. [q1 V(r1) + q2 V(r2) + q1q2/4πε0 (r1 – r2)]

Ans3 (positive , negative)

Ans4- electrostatic shielding. Explain.

Ans5- (3 V)

Solution- 1/CS= 1/6 + 1/12 i.e. Cs = 4 μF.

charge across 6μF= CV= 6x2= 12μC. It is equal to total charge in the circuit as both are connected in series.

Voltage across 12μF= q/c= 12/12= 1 V

Total supply voltage= 2+1= 3 V.

Ans 6- (on connecting potential will be same. So q1/ 4πε0R1 = q2/4πε0R2 so q1/q2 = R1/R2. And the charge density ratio= σ1:σ2= q1/4πR1

2: q2/4πR22 so ratio of their charge densities R2/R1)

Ans7. Let a point inbetween at the distance x from 4 Q then find electric field intensity due to both of the charges. On solving X=66.7 m or x=2 m where x is the distance of point p from charge 4Q where electric field intensity is zero.

Electrostatic potential energy U=1/4π ε0 . q1q2/r= 41.44x10-3 J

Ans. 8:-

Ans9. Electric flux linked with any surface is the total number of electric field lines passing through the surface normally.

Its SI unit is Nm2C-1 or JmC-1 or Vm.

Remain unchanged.

Ans10.

Ans11:

Ans12. Cs=9μF as they are connected in series.

When connected in parallel Cp= 27 μF

Since voltage is same, so Energy is directly proportional to capacitance

Es/Cs= Ep/Cp

So, Es/Ep = Cs/Cp

Es:Ep= 1:3

Ans13.

Ans14. (i) since electric field is conservative in nature, the amount of work done depends on initial and final position.

So work done W=F.d = q E.d = -qE4cos 1800= +4qE

Hence VC-VA = W/q

Hence VC-VA = W/q = 4E

Ans15.

(i) The dipole moment of dipole is P= qx (2a)Force on –q at A = -qEforce on +q at B = +qEnet force on the dipole = qE –qE = 0

(ii) Work done on dipoleW = ∆U = PE( cosɵ1 - cosɵ2) = PE( cos0o – cos180o) W = 2PE

Ans16. We know

Capacitance with dielectric of thickness ‘t’

C = ɛoA d-t+t/K

After putting the value of t=d/2 the value of C will be

C= 2ɛoAK d(K+1)

Ans 17. Let q be the charge on the charged capacitor.

So, energy stored in it is given by U= q2/2CWhen another uncharged similar capacitor is connected, then the net capacitance of the system is given by C’= 2C The charge on the system remains constant. So, the energy stored in the system is given by U’= q2/2C’ = q2/4C as C’= 2CThus the required ratio is given by U’/U= ½

Ans18. Let an electric dipole be rotated in electric field be rotated in electric field from angle ɵo to ɵ1 in the direction of electric field.

Torque on dipole = PEsinɵ

The work done i rotating the dipole further through a small angle dɵ is

dW = PEsinɵdɵ

total work done in rotating the dipole from angle ɵo to ɵ1 is given by integrating the expression within the limits and the result obtained is given by

w = PE(cosɵo - cosɵ1)

THREE MARKS QUESTIONS

1. The battery remains connected to a parallel plate capacitor and a dielectric slab is inserted between plates. What will be the effect on its capacitance, charge, potential difference, electric field, energy stored? Justify your answer.

2. The battery connected to a parallel plate capacitor is removed and a dielectric slab is inserted between plates. What will be the effect on its capacitance, charge, potential difference, electric field, energy stored? Justify your answer. Where does the loss of energy stored go?

3. A spherical conducting shell of inner radius r1 and outer radius r2 has a charge Q. (i) a charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell? (ii) is the electric field intensity inside a cavity with no charge is zero, even if the shell is not spherical? Explain.

4. Define electric flux. Is it a scalar or a vector quantity? A point charges q is at a distance of d/2 directly above the centre of a square of side d. Use Gauss' law to obtain the expression for the electric flux through the square. (b) If the point charge is now moved to a distance 'd' from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.

5. A positive point charge (+q) is kept in the vicinity of an uncharged conducting plate. Sketch the electric field lines originating from the point on to surface of the plate.

Derive expression for the electric field at the surface of a charged conductor.6. Use Gausses’ law to derive the expression for the electric field between two uniformly charged parallel

sheets with surface charge densities σ and –σ respectively.7. A sphereS1 of radius r1 encloses a net charge Q. If there is another concentric sphere S2 of radius r2 (r2˃

r1) enclosing charge 2Q.(i) Find the ratio of the electric flux through sphere S1 and S2.(ii) How will the electric flux through sphere S1 change, if a medium of dielectric constant 5 is

introduced in the space inside S1 in place of air?

8. Two capacitors of unknown capacitances C1 and C2 are connected first in series and then in parallel across a 100 V battery. If the energy stored in the two combinations is 0.045J and 0.25 J respectively, then determine the value of C1 andC2. Also calculate the charge on each capacitor in parallel combination.

9.

A thin metallic spherical shell of radius R carries a charge Q on its surface. A point charge Q/2 is placed at the centre C and another charge +2Q is placed outside the shell at A at a distance x from the centre as shown in the figure.

(i) Find the electric flux through the shell(ii) State the law used(iii) Find the force on the charges at the centre C of the shell and at the point A.

10. Two identical parallel plate capacitors A and B are connected to a battery of V volts with switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric.

11. Four point charges Q, q, Q and q are placed at the corners of a square of side ‘a’ as shown in the figure. Q q

q Q

Find the

a) Resultant electric force on a charge Q, andb) Potential energy of this system.

12. a. Three point charges q, -4q and 2q are placed at the vertices of an equilateral triangle ABC of side ‘ l’ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge q.

q A

-4q B C 2q

Find out the amount of work done to separate the charges at infinite distance.13. Calculate the equivalent capacitance between points A and B in the circuit below. If a battery of 10 V is

connected across A and B, calculate the charge drawn from the battery by the circuit.

ANSWERS OF QUESTIONS CARRYING 3 MARKS

Ans1. As the battery is connected so p.d. will be same across the plates of the capacitor . V= same, C0= ε0A/d but on inserting the slab C= K ε0A/d so C= KC0, Q= CV= KQ0, E will be same. U= ½ CV2= KU0

Ans2- As battery is removed so charge will remain same as the source of charge has been removed. Explain further.

- (q same, kC0, V0/k, E0/k, U0/k). in this process dielectric slab is polarized. The energy lost is spent in polarizing the slab)

Ans3- ( -q/4πr12, Q+q/4πr2

2,0 ) (when charge q is placed at the centre of the shell, a charge –q is induced at the inner surface and +q at the outer surface of the shell)

Answer 4:- (a) Electric flux is defined as the number of electric filed lines crossing the per unit area. It is scalar quantity. When cube is of side d and point charge q is at the center of the cube then the total electric flux due to this charge will pass evenly through the six faces of the cube. So, the electric flux through one face will be equal to 1/6 of the total electric flux due to this charge. i.e., Flux through 6 faces = q/ 0 Flux through 1 face = 1/6 x q/ 0 (b) If we moved point charge d from centre and square side changes to 2d, still the point charge can be imagined at the center of a cube of side 2d. Again the flux through one face of the cube will be 1/6 of the total electric flux due to the charge q. Hence, the electric flux through the square will not change and it will remain the same i.e., q/6∈o.

Ans 5-

For 2nd part see NCERT Text Book page No. 69.

Ans 6:-Let us consider two uniformly charged parallel sheets carrying charge densities +σ and –σ respectively, separated by a distance r

By Gauss’ s law it is known that electric field due to a uniformly charged infinite plane sheet at any nearby point is E=σ/2ε0.

The electric field is directed normally outward from the plane sheet for positive charge and normally inward for negative charge.

Let r represents unit vector directed from positive plate to negative plete.

Now electric field intensity at any point P between the plates is given by

(i) E1=+σ/ 2ε0r (due to positive charge)(ii) E2=- σ/ 2ε0r (due to negative charge)

So , electric field intensity at point P is given by E= E1+ E2 = σ/ 2ε0r + σ/ 2ε0r = σ/ ε0r

Ans7. (i) Ф1 = Q/εo, Ф2= 3Q/εo Ф1/Ф2 =1/3

ii) if a dielectric medium of dielectric constant 5 is inserted then

Ф1’ =Q/5εo= Ф1/5

Ans 8. Energy stored in series combination is

Es = 1/2CsV2 =C1C2x(100)2/2(C1+C2) =0.045J ----------(1)

Energy stored in parallel combination is

Ep= ½ CpV2 =1/2 (C1+C2)V2 = 0.25 J-----------------------(2)

On solving equations 1 and 2 we get

C1= 35µF and C2 =15µF

Q1 = 35x10-4C and Q2 =15x10-4C

Ans9. (i) total charge enclosed = Q+Q/2 = 3Q/2

So, Ф = 3Q/2ε0

ii) Statement of Gauss’s law along with mathematical expression

(iii) Force on charge at A, F = 3Q2/4πε0x2

Ans10. Two capacitors are connected in parallel. Hence, the potential on each of them remains the same. So the charge on each capacitor is

QA =QB = CV

Net capacitance with switch S closed = C+C =2C so, energy stored = 1/2X2CXV2 =CV2

After the switch S is opened, capacitance of each capacitor= KC

In this case, voltage only across A remains the same.

The voltage across B changes to V’=Q/C=Q/KC

Energy stored in capacitor A= 1/2KCV2

Energy stored in capacitor B= Q2/2KC

Total energy stored=1/2KCV2+ Q2/2KC= CV2[K2+1]/2K

Required ratio= 2K/(K2+1)

Ans11. The resultant force on the charge Q is given by

FQ = 1/4πɛoa(2Qq)1/2 Potential energy of the system = 1/4πɛ₀[qQ/a+QQ/a√2+qQ/a+qQ/a√2+qQ/a+qQ/a] = 1/4πɛ₀[3qQ/a+(Q2+qQ/a√2)]

Ans 12.a. The magnitude of the resultant electric force acting on the charge q is given by

Fq=1/4πɛ₀[(-4q)(q)/l2+(2q)(q)/l2]

= -2q2/4πɛ₀l2

b. The work done to separate the charges at infinite distance = -potential energy of the system

=1/4πɛ₀[(-4q)(q)/l+(2q)(q)/l+(-4q)(2q)/l]

= +[10q2/4πɛ₀]

Ans13. Since C1 /C2 = C3/C4

This is the condition of balance so there will be no current across PR

Now C1 and C2 are in series,

So C12 = 20/3 µF

again C3 and C4 are in series

So, C34 = 10/3µF

Equivalent capacitance between A and B is CAB = C12+ C34 = 10 µF.

Charge drawn from battery Q = CV = 10x10µC = 100µC

FIVE MARKS QUESTIONS

1. Derive an expression for the energy stored in a parallel plate capacitor.On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved, and a dielectric medium of ɛr =10 is introduced between the plates, without disconnecting the d.c. source. Explain using suitable expression, how the (i) capacitance, (ii) electric field and (iii) energy stored in the capacitor change.

2. (a) Define electric flux. Write its SI unit.(b) The electric field components due to a charge inside the cube of side 0.1 m . Ex = αx where α = 500 N/Cm

Ey = 0, Ez = 0Calculate (i) the flux through the cube, and (ii) the charge inside the cube.

3. a. Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.b. Draw the equipotential surfaces due to an electric dipole. Locate the points where the potential due to the dipole is zero.

4. Using Gauss’ s law deduce the expression for the electric field due to a uniformly charged spherical conducting shell of radius R at a point (i) outside and (ii) inside the shell. Plot a graph showing the variation of electric field as a function of r ˃ R and r ˂ R (r being the distance from the centre of the shell).

5. a. Deduce the expression for the torque acting on a dipole of dipole moment P in the presence of a uniform electric field.

b. Consider two hollow concentric spheres, S1 and S2, enclosing charges 2Q and 4Q respectively as shown in figure.

S2 4

4Q S1

2Q

(i) find out the ratio of electric flux through them.(ii) How will the electric flux through the space S1 change if a medium of dielectric constant ɛr is introduced in the space inside S1 in place of air? Deduce the necessary expression.

6. a. Deduce the expression for the potential energy of an electric dipole of dipole moment P placed in a uniform electric field E.

Find out orientation of the dipole when it is in (i) stable equilibrium (ii) unstable equilibrium.

b. Figure shows a configuration of the charge array of two dipoles.

a a +q------------------------------ -2q------------------------------+q…………………….p

r

Obtain the expression for the dependence of potential r for r˃>a for a point p on the axis of this array of charges.

7. a. Define electric flux. Write its S.I unit.

b.Using Gauss’ Law, obtain the electric flux due to a point charge ‘q’ enclosed in a cube of side ‘a’.

c. Show that the electric field due to a uniformly charged plane sheet at any point distant x from it, is independent of x.

8. a. Derive an expression for the electric field E due to a dipole of length ’2a’ at a point distant r from the centre of the dipole on the axial line.

b. Draw a graph of E versus r for r>>a.

c. If this dipole were kept in a uniform external field E0, diagrammatically represent the position of the dipole in stable and unstable equilibrium and write the expressions for the torque acting on the dipole in both the cases.

9. a. Use Gauss’ theorem to find the electric field due to a uniformly charged infinitely large plane thin sheet with surface charge density ơ.

b. An infinitely large thin plane sheet has a uniform surface charge density + ơ. Obtain the expression for the amount of work done in bringing a point charge q from infinity to a point, distant r, in front of the charged plane sheet.

10.a) Define electric flux. Is it a scalar or a vector quantity?

A point charge ‘q’ is at a distance of d/2 directly above the centre of a square of side d, as shown in the figure. Use Gauss’ Law to obtain the expression for the electric flux through the square.

q

d/2 d

b) If the point charge is now moved to a distance ‘d’ from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected?

11. Use Gauss’ Law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density λ C/m.

a) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.b) Find the work done in bringing a charge q from perpendicular distance r1 to r2 (r2>r1)

ANSWERS OF FIVE QUESTIONS

Ans1. For energy stored, see the NCERT Text Book.

When capacitor is charged by a battery and remains connected, the potential difference remains same, equal to V.

(i) C =ɛoA/d d’ = d/2 and ɛ r = 10So, C’ = ɛrɛoA/d/2 = 2ɛr ɛoA/d = 2x10C = 20C

(ii) E =V/d E’ = V/d/2 = 2 V/d = 2 E(iii) Energy density initially, U = ½ ɛoE2

Energy density finally , U’= ½ ɛrɛoE’2 = 40 U

Ans 2. The number of electric field lines passing through an area normally is called electric flux.

Its SI unit is Nm2C-1 or JmC-1 or Vm.

a) See NCERT text book (page no. 35)

Ans3. It is defined as the product of either charge and the length of the dipole.

Mathematically, P= q (2a).

It is a vector quantity.

For derivation see NCERT Text Book page no. 28.

b). for drawing see NCERT Text Book page no. 60. Ans 4. For derivation see NCERT Text Book page no. 39. Graph:

Ans5. Consider an electric dipole of charge –q and +q and of length 2a placed in a uniform electric field E making an angle ɵ with electric field.

Force on charge –q = -qE force on charge +q = +qE electric dipole is under the action of two equal and opposite unlike parallel forces, which give rise to a torque on the dipole.

N B +qE

ɵ

-qE A

Ϯ = qE(AN)

= q(2a)Esinɵ = PEsinɵ

Ϯ = PXE

b. (i) Charge enclosed by sphereS1 = 2Q

Electric flux through the sphereS1 is, Ф= 2Q/ɛ₀

Charge enclosed by sphereS2 =2Q+4Q = 6Q

Electric flux through the sphereS21 is, Ф1= 6Q/ɛ₀

Ratio is given by Ф/Ф1 = 1:3

(ii) for sphere S1, the electric flux is Ф’ = 2Q/ɛr

Ф’/Ф1= 2Q/ɛr/6Q/ɛ₀ = ɛr/ɛ₀.1/3

Since ɛr>ɛ₀ , Ф’< Ф1

Therefore the electric flux through the sphere S1 decreases with the introduction of the dielectric medium.

Ans6.a. Let us suppose that the electric dipole is brought from infinity in the region of a uniform electric field such that the electric dipole moment P always remains perpendicular to electric field. The electric forces on charges +q and -q are +qE and –qE, along the field direction and opposite to the field direction respectively.

As charges +q and –q traverse equal distance under equal and opposite forces; therefore net work done (W1) in bringing the dipole in the region of electric field perpendicular to the field direction will be zero.

Now, the dipole is rotated and brought to orientation making an angle ɵ with the field direction

(ɵ₀ = 90₀ and ɵ1 =ɵo), therefore work done

W2 = PE( cos ɵ₀- cos ɵ1) = -PEcosɵ

So total work done in bringing the dipole from infinity . i.e. electric potential energy of the dipole

U = W1+W2 = 0- PEcosɵ = -PEcosɵ

In vector form U = - P.E

When ɵ=0o, i.e. the dipole is along the direction of the electric field corresponds to stable equilibrium.

When ɵ =1800 i.e. the dipole is anti parallel to the direction of the electric field corresponds to unstable equilibrium.

b. potential at the point P is given by

V = 1/4πɛ₀[q/(r+a) +(-2q)/r +q/r-a]

= q/4πɛ₀[1/(r+a) -2/r+ 1/(r-a)]

= q/4πɛ₀[r2+a2/r(r2-a2)]

As r>> a , V = q/4πɛ₀r

Ans 7: (a) The number of electric field lines passing through an area normally is called electric flux.

Its SI unit is Nm2C-1 or JmC-1 or Vm.

b. According to Gauss’ slaw states that the total electric flux through a closed surface is equal to 1/ε0 times the magnitude of the charge. i.e. Ф =q/ε0

Here the charge enclosed by the cube is q , so electric flux through the cube is given by Ф =q/ε0

c. See NCERT Text Bok page no. 38.

Ans8. a. See NCERT Text Book page no.27.

b.

E

r

C. -q +q

+q -q

Ans 9. a.See NCERT Text Book page no. 38.

b.

Ans10. a) The number of electric field lines passing through an area normally is called electric flux.

It is a scalar quantity.

Here the given square of side d is one face of the cube of side a. At the centre of the cube a charge q is placed.

According to Gauss’ theorem the total electric flux through the six faces of the cube = q/εo

So the total electric flux through the square = q/6ε0

b.If the point charge is moved to a distance ‘d’ from the centre of the square and the side of the square is doubled then there will be no change in the electric flux . As here also the charge is at the centreof the cube.

Ans 11. a)

b.

E

r

.