Class 12 Physics MCQ Assignment

Distance from +4μC charge = 4m

Distance from −2μC charge = 5m

V = k[(4×10⁻⁶)/4 − (2×10⁻⁶)/5]

V = 9×10⁹ × 0.6×10⁻⁶

V = 5.4 × 10³ V

Potential at center of ring:

V = kQ/R

Work done = qV

Therefore,
W = kQq/R

Inside a conductor in electrostatic equilibrium, electric field is zero.

Hence potential remains constant throughout the conductor.

E = −∇V

Ex = −∂V/∂x = −10xy
Ey = −∂V/∂y = −5x²

Therefore,
E = (−10xy, −5x², 0)

C = ε₀A/d

C = (8.85×10⁻¹² × 0.02)/(2×10⁻³)

C = 88.5 × 10⁻¹² F

C = 88.5 pF

For isolated capacitor:

U = Q² / 2C

Since capacitance is inversely proportional to plate separation:

C ∝ 1/d

When separation doubles, capacitance becomes half.

Therefore energy doubles.

Potential due to electric dipole at axial point:

V = kp / r²

Hence potential varies inversely as square of distance.

On an equipotential surface:

Potential difference = 0

Therefore work done:

W = qΔV = 0

Initial charge:

Q = CV

Q = 4 × 100 = 400 μC

Total capacitance after connection:

C = 4 + 6 = 10 μF

Final voltage:

V = Q/C = 400/10 = 40V

Potential due to infinite line charge:

V ∝ ln(r)

Hence potential varies logarithmically with distance.

Total electrostatic energy equals sum of interaction energies of all pairs.

There are:
4 side pairs → distance = a
2 diagonal pairs → distance = √2 a

U = 4(kq²/a) + 2(kq²/√2a)

U = 4kq²/a + √2 kq²/a

U = (4 + √2)kq²/a

Capacitance with dielectric:

C' = KC

Here K = 5

Therefore:

C' = 5C

Work done:

W = qV

W = 5 × 10⁻⁶ × 120

W = 6 × 10⁻⁴ J

W = 0.6 mJ

Potential of sphere:

V = kQ/R

Electrostatic energy:

U = ½QV

U = ½ × Q × (kQ/R)

U = kQ²/2R

Uniform electric field means:

Electric field strength remains constant with distance.

Therefore E vs distance graph is a horizontal straight line.

Every charge element lies at same distance R from center.

Potential:

V = kQ/R

Electric potential is a scalar quantity.

It has no direction and follows algebraic superposition.

Charge stored:

Q = CV

Q = 5 × 10

Q = 50 μC

Energy formula:

U = ½CV²

18 = ½ × C × 36

C = 1F

All points on an equipotential surface have same potential.

Therefore:

ΔV = 0

Kinetic energy gained:

KE = eV

Therefore:

KE = 500 eV

Dielectric becomes polarized in external electric field.

Induced field opposes applied electric field.

Hence net electric field decreases.

Potential of sphere:

V ∝ Q

If potential becomes 3V:

Final charge = 3Q

Additional charge:

3Q − Q = 2Q

Electrostatic potential energy:

U = kq₁q₂/r

U = (9×10⁹ × 2×10⁻⁶ × −3×10⁻⁶)/0.2

U = −0.27 J

Energy stored:

U = ½CV²

U = ½ × 2×10⁻⁶ × (200)²

U = 0.04 J

Potential of sphere:

V = kQ/R

If radius becomes half:

V' = 2V

Therefore:

V' = 2 × 6 × 10⁸

V' = 1.2 × 10⁹ V

Electric field:

E = −dV/dr

Equal spacing means constant potential gradient.

Hence electric field is uniform.

Since capacitor remains connected to battery:

Battery maintains constant voltage.

Therefore potential difference remains constant.

Capacitance of isolated sphere:

C = 4πε₀R

C = 4π × 8.85×10⁻¹² × 2

C ≈ 222 pF

Potential energy:

U = qV

For positive charge, decrease in potential means decrease in potential energy.

Electric field is related to potential by:

E = −dV/dx

Hence slope of V-x graph gives negative electric field.

In electrostatic equilibrium:

Electric field inside conductor must be zero.

Otherwise free electrons would move continuously.

Potential energy of dipole:

U = −pE cosθ

At θ = 60°:

U = −pE × 1/2

U = −pE/2

Parallel combination:

2C + C = 3C

Now two 3C capacitors are in series:

Ceq = (3C × 3C)/(3C + 3C)

Ceq = 3C/2

Electric field:

E = −dV/dr

If potential is constant:

dV/dr = 0

Therefore electric field is zero.

Electric potential:

V = W/q

SI unit:

Volt = Joule/Coulomb

Therefore unit is J/C.

Potential due to point charge:

V = kQ/r

Therefore:

V ∝ 1/r

Energy density:

u = ½ ε₀E²

Hence energy density is proportional to square of electric field.

SI unit of capacitance is:

Farad (F)

In parallel combination:

Potential difference across each capacitor remains same.

Therefore all capacitors have same voltage.

In series combination:

Same charge flows through each capacitor.

Hence all capacitors carry equal charge.

In electrostatics:

Potential at infinity is chosen as reference potential.

Therefore:

V∞ = 0

Energy stored:

U = Q²/2C

Therefore:

U ∝ Q²

If charge doubles:

U becomes four times.

Electric field direction is always normal to equipotential surface.

Therefore field lines are perpendicular to equipotential surfaces.

Capacitance depends on:

• Shape and size of conductors
• Distance between conductors
• Dielectric medium

Hence all options are correct.

Electrostatic field is conservative in nature.

Therefore work done over any closed path:

W = 0

Electric field inside conducting shell is zero.

Hence potential remains constant throughout interior region.

For two equal capacitors:

1/Ceq = 1/C + 1/C

1/Ceq = 2/C

Ceq = C/2

In parallel combination:

Ceq = C + C

Ceq = 2C

Electric field:

E = −dV/dr

Smaller spacing between equipotential surfaces means larger potential gradient.

Hence electric field becomes maximum.