CBSE Class 12 Physics Electrostatics important questions

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Worksheet 1-A Name of the chapter: Electrostatics Name of the Topic:

ELECTROSTATICS IMPORTANT QUESTIONS
CLASS XII : WORKSHEETS LEVEL A, B AND C

WORKSHEET 1-A (Level A)

1. Two equal balls having equal positive charge ‘q’ coulombs are suspended by two insulating strings of equal length .What would be the effect on the force when a plastic sheet is inserted between the two?

Answer:
1. Force between two point charges in air: F = q²/(4πε₀r²).
2. When a dielectric of constant K fills the space between the charges, the force becomes F' = F/K.
3. Hence the force reduces to 1/K times its original value.

Teacher's Note:
a) The dielectric weakens the electric force between the charges by the factor K.
b) The insulating strings do not change the electrostatic force itself; only the medium changes.

2. An electric dipole is held in a uniform electric field .What is the net force acting on it?

Answer:
1. In a uniform field the forces on the two charges of the dipole are equal in magnitude and opposite in direction.
2. Net force = qE + (−qE) = 0.
3. Hence the net force on the dipole is zero, although a torque may act on it.

Teacher's Note:
a) A uniform field produces no net force on a dipole; it produces only a torque.
b) Do not confuse net force with net torque, which is not zero when the dipole is inclined to the field.

3. Net capacitance of three identical capacitors in series is 1 µF. What will be their net capacitance if connected in parallel?

Answer:
1. Series: 1/C_s = 1/C + 1/C + 1/C = 3/C, so C_s = C/3.
2. Given C_s = 1 µF, so C = 3 µF for each capacitor.
3. Parallel: C_p = C + C + C = 3C = 9 µF.
4. Hence the net capacitance in parallel is 9 µF.

Teacher's Note:
a) In series the reciprocals add, so the net value is smaller than each capacitor.
b) In parallel the capacitances add directly, so find C first from the series value.

4. How will the (i) energy stored and (ii) the electric field inside the capacitance be affected when it is completely filled with a dielectric material of dielectric constant ‘K’.

Answer:
1. The question does not state whether the capacitor remains connected to the battery. The two cases are given below.
2. Case 1, capacitor isolated (charge constant): energy becomes U/K and the electric field inside becomes E/K.
3. Case 2, capacitor kept connected to the battery (voltage constant): the capacitance becomes KC, so the energy becomes KU, and the electric field inside stays the same, since E = V/d.

Teacher's Note:
a) The printed key says that energy increases K times and the field reduces K times; that pairing is not valid for either case.
b) With a battery connected the field stays unchanged, because the potential difference and the plate separation are both fixed.
c) For an isolated capacitor, the induced charge on the dielectric reduces the field by K, and the energy falls by the same factor.

5. Two charges of magnitude -2Q and +Q are located at point (a, 0) and (4a, 0) respectively . What is the electric flux due to these charges through a sphere of radius ‘4a’ with its centre at the origin?

Answer:
1. The charge −2Q at (a, 0) lies inside the sphere, since its distance from the origin is a, which is less than 4a.
2. The charge +Q at (4a, 0) lies on the surface of the sphere, and it is counted as enclosed in this worksheet.
3. Net enclosed charge = −2Q + Q = −Q.
4. By Gauss's law, flux φ = (net enclosed charge)/ε₀ = −Q/ε₀.

Teacher's Note:
a) Find the distance of each charge from the origin before deciding whether it is inside.
b) A charge on the surface is a boundary case; the worksheet counts it as enclosed, and the answer −Q/ε₀ follows from that convention.
c) Flux is independent of the position of the charges inside the sphere, only their total charge matters.

WORKSHEET 1-B (Level B)

1. A charge ‘q’ is placed at the centre of a cube of side l. What is the electric flux passing through each face of the cube?

Answer:
1. Total flux through the closed cube = q/ε₀, because the charge is inside.
2. The six faces are identical and symmetric about the centre.
3. Flux through each face = (q/ε₀)/6 = q/(6ε₀).

Teacher's Note:
a) The charge at the centre sees every face in the same way, so the flux divides equally.
b) The side length l does not affect the answer.

2. An electric dipole is held in a uniform electric field , Write the expression for torque acting on it?

Answer:
1. Torque on a dipole in a uniform field is τ = p × E.
2. Magnitude: τ = pE sinθ, where θ is the angle between the dipole moment p and the field E.

Teacher's Note:
a) The torque is zero when θ is 0° or 180°, and maximum when θ is 90°.
b) The direction of the torque is perpendicular to both p and E.

3. Net capacitance of three identical capacitors in series is 1 µF. Now they are connected in parallel ,Find the ratio of energy stored in the two configurations if they are both connected to the same source.

Answer:
1. As in Worksheet 1-A, each capacitor is 3 µF, the series value is 1 µF and the parallel value is 9 µF.
2. Energy stored in a configuration = ½ C V², with the same source voltage V.
3. Series energy : parallel energy = 1 : 9.
4. Hence the ratio of energy stored (series : parallel) is 1 : 9.

Teacher's Note:
a) Energy is proportional to capacitance when the voltage is the same.
b) The parallel arrangement stores nine times the energy of the series arrangement at the same source voltage.

4. A spherical conducting shell of inner radius r1 and r2 has a charge ‘Q’ .A charge ‘q’ is placed at the centre of the shell. What is the surface charge density on the (i) inner surface.(ii) outer surface of the shell?

Answer:
1. Here r1 is the inner radius and r2 is the outer radius of the shell. The shell carries total charge Q.
2. The charge q at the centre induces a charge −q on the inner surface, which is spread uniformly.
3. Surface charge density on the inner surface: σ_inner = −q/(4πr₁²).
4. The outer surface carries the remaining charge, Q + q, so σ_outer = (Q + q)/(4πr₂²).

Teacher's Note:
a) The inner surface charge is always equal and opposite to the enclosed charge q.
b) The outer surface carries the total charge of the shell plus the induced charge.
c) Use the surface area 4πr² of each surface in the density formula.

5. Two charges of magnitude -Q and +Q are located at point (a, 0) and (4a, 0) respectively . What is the electric flux due to these charges through a sphere of radius ‘4a’ with its centre at the origin?

Answer:
1. The charge −Q at (a, 0) is inside the sphere of radius 4a.
2. The charge +Q at (4a, 0) lies on the surface, and it is counted as enclosed here.
3. Net enclosed charge = −Q + Q = 0.
4. Flux φ = 0/ε₀ = 0.

Teacher's Note:
a) This is the same geometry as Worksheet 1-A, Question 5, with the magnitude of the first charge changed to Q.
b) The net enclosed charge is zero, so the flux is zero even though the field is not zero on the surface.

WORKSHEET 1-C (Level C)

1. Define electric dipole moment. Write its SI unit .

Answer:
1. Electric dipole moment is a vector quantity equal to the product of the magnitude of either charge and the separation between the charges.
2. It is written as p = q × d, where d is the separation directed from the negative to the positive charge.
3. SI unit: coulomb metre (C m).

Teacher's Note:
a) The printed key writes P = 2q.l, which is not a correct definition; the correct expression is p = q × d.
b) The separation d is the full distance between the two charges, not half of it.
c) The dipole moment points from the negative charge towards the positive charge.

2. An electric dipole isalligned parallel to a uniform electric field. Find the workdone in rotating it through the angle of 180°.

Answer:
1. Initially the dipole is parallel to the field, so θ₁ = 0°.
2. Work done in rotating a dipole from θ₁ to θ₂ is W = pE(cosθ₁ − cosθ₂).
3. Here θ₂ = 180°, so W = pE(cos 0° − cos 180°) = pE(1 − (−1)) = 2pE.
4. Hence the work done is 2pE.

Teacher's Note:
a) The work done equals the change in potential energy, which is U = −pE cosθ.
b) At 180° the dipole is antiparallel and has the maximum potential energy 2pE from its starting value.
c) The work is positive because an external agent must supply energy to turn the dipole.

3. Name the physical quantity whose S.I unit is JC-1 .Is it scalar or a vector quantity?

Answer:
1. Joule per coulomb (J C⁻¹) is the unit of electric potential.
2. Electric potential is a scalar quantity.

Teacher's Note:
a) Electric potential is energy per unit charge, so its unit is J C⁻¹, or volt.
b) Electric field is a vector quantity, while its unit is N C⁻¹; do not confuse the two.

4. Two charges of magnitude -2Q and +Q are located at point (a, 0) and (4a, 0) respectively . What is the electric flux due to these charges through a sphere of radius ‘3a’ with its centre at the origin?

Answer:
1. The charge −2Q at (a, 0) is at distance a from the origin, which is less than 3a, so it is inside.
2. The charge +Q at (4a, 0) is at distance 4a, which is greater than 3a, so it is outside.
3. Net enclosed charge = −2Q.
4. Flux φ = −2Q/ε₀.

Teacher's Note:
a) Only charges inside the Gaussian surface contribute to the flux.
b) The external charge +Q contributes to the field on the surface but not to the net flux.

5. A spherical shell of radius 2 cm has a charge 10 µC at its centre. What is the work done in bringing a 5 µC charge from one point on the surface to diametrically opposite point on the surface?

Answer:
1. Every point on the surface of the shell is at the same potential, V = kQ/R.
2. Both diametrically opposite points are on the surface, so their potentials are equal.
3. Work done = q(V_B − V_A) = 5 µC × 0 = 0.
4. Hence the work done is zero.

Teacher's Note:
a) Work done in moving a charge between two points equals the potential difference times the charge.
b) Equal potential on a surface means no work is needed to move a charge along it.
c) The radius 2 cm and the charge values do not affect the answer.

WORKSHEET 1-A NAME OF THE CHAPTER: Electrostatics Name of the Topic:

ANSWERS

1. Force reduces

2. Zero

3. 9μF

5. ᶲ = _-Q/€0 

Worksheet 1-BName of the chapter: Electrostatics

Answer Key

WORKSHEET 1-B NAME OF THE CHAPTER: Electrostatics Name of the Topic:

ANSWERS

1. ᶲ = _q/6€0

2. τ = pE sinϴ

3. 1: 9

4. (i) –q/4πr¹

WORKSHEET 1-C NAME OF THE CHAPTER: : Electrostatics

Name of the Topic:

ANSWERS

1. THE Strength of a dipole is measured by a vector quantity known as dipole moment.It is product of either charge and dipole length. P= 2q.l , S.I Unit is Coulomb meter

2. W = 2PE

3. Electric Potential, Scalar quantity

4. ᶲ = _-2Q/€0

5. Zero