Capacitors - Free MCQ Practice Test with solutions, JEE Physics

Q = CV, where C is decided only by the geometrical factors

The electric field between the plates of a capacitor can be calculated using the formula:

• Electric Field (E) = Potential Difference (V) / Distance (d)

Given:

• Potential Difference (V) = 150 V
• Distance (d) = 6 mm = 0.006 m (conversion from mm to m)

Calculation:

• E = 150 V / 0.006 m
• E = 25,000 V/m or 2.5 × 10⁴ V/m

Thus, the electric field at points A and B is 2.5 × 10⁴ V/m.

Alcohol and HCl are polar molecules since they have a net dipole moment towards a particular direction. Both water and benzene are non-polar molecules. But water is a conductor of electricity, whereas benzene is a dielectric (insulator).

Given:

• Original distance between plates = d

• Dielectric medium = air (or vacuum)

• Original capacitance (C0) = (epsilon0 * A) / d
  where A is plate area, and epsilon0 is the permittivity of free space.

Effect of Introducing a Metal Plate:

When a metal plate of thickness d/2 is inserted between the plates without touching them, it splits the capacitor into two capacitors in series:

1. First capacitor: between original plate and metal plate (separation d1)

2. Second capacitor: between metal plate and other plate (separation d2)

Since the metal plate is thin and placed symmetrically:
d1 + d2 = d - t = d - d/2 = d/2

If the metal plate is exactly in the middle, then:
d1 = d2 = d/4

New Capacitance Calculation:

Each of the two capacitors has capacitance:
C1 = C2 = (epsilon0 * A) / d1 = (epsilon0 * A) / (d/4) = (4 * epsilon0 * A) / d

When connected in series, the equivalent capacitance (Ceq) is:
1 / Ceq = 1 / C1 + 1 / C2
=> 1 / Ceq = d / (4 * epsilon0 * A) + d / (4 * epsilon0 * A) = d / (2 * epsilon0 * A)
So, Ceq = (2 * epsilon0 * A) / d

Comparison with Original Capacitance:

Original capacitance (C0) = (epsilon0 * A) / d
New capacitance (Ceq) = (2 * epsilon0 * A) / d = 2 * C0

Conclusion:
The capacitance doubles when a metal plate of thickness d/2 is introduced.

Answer: (b) It will be double of its initial value.

C=(Kε0A/d)∝K/d
Hence,C₁/C₂=d2/d1=(d/3)d=3
C₂=3C₁

Capacitance of an isolated spherical conductor =4π€R , where €=permittivity of vacuum
C=2×10^-6F
C=4π€R
4×3.14×8.85×10^-12×R=2×10^-6
R=2×10^-6/4×3.14×8.85×10^-12
=1.8×10⁴ m

Given, charge =q=8×10−18C and Capacitance =C=100μF
Here the work done is the energy stored in the capacitor. i.e,
W=q^2​/C= (8×10−18)²/100×10⁻⁶​=32×10⁻³²J

Mica capacitors are preferred for high frequency circuits because they have low ohmic losses and less reactance.

Q = CV

W = Energy stored in the capacitor