Question

A parallel-plate capacitor of capacitance 40µF is connected to a 100 V power supply. Now the intermediate space between the plates is filled with a dielectric material of dielectric constant K = 2. Due to the introduction of dielectric material, the extra charge and the change in the electrostatic energy in the capacitor, respectively, are -

• A. 2 mC and 0.2 J
• B. 8 mC and 2.0 J
• C. 4 mC and 0.2 J
• D. 2 mC and 0.4 J

Hint:

When a dielectric is inserted into a capacitor connected to a power supply, the charge increases and the energy stored also increases.

Answer 1

The Correct Option is C

Step 1: Determine the additional charge.

The difference in charge (\(\Delta q\)) introduced by the dielectric is calculated as:

\(\Delta q = (KC - C)V\)

Where K represents the dielectric constant, C signifies the capacitance, and V denotes the voltage.

\(\Delta q = (2 \times 40 \times 10^{-6} \text{ F} - 40 \times 10^{-6} \text{ F}) \times 100 \text{ V}\)

\(\Delta q = (80 - 40) \times 10^{-6} \text{ F} \times 100 \text{ V}\)

\(\Delta q = 40 \times 10^{-6} \text{ F} \times 100 \text{ V}\)

\(\Delta q = 4000 \times 10^{-6} \text{ C} = 4 \times 10^{-3} \text{ C} = 4 \text{ mC}\)

Step 2: Calculate the alteration in electrostatic energy.

The change in electrostatic energy (\(\Delta U\)) is computed using the formula:

\(\Delta U = \frac{1}{2} C' V^2 - \frac{1}{2} C V^2 = \frac{1}{2} (KC - C) V^2 = \frac{1}{2} (K - 1) C V^2\)

\(\Delta U = \frac{1}{2} (2 - 1) (40 \times 10^{-6} \text{ F}) (100 \text{ V})^2\)

\(\Delta U = \frac{1}{2} (1) (40 \times 10^{-6} \text{ F}) (10000 \text{ V}^2)\)

\(\Delta U = \frac{1}{2} (40 \times 10^{-2}) \text{ J}\) \(\Delta U = 20 \times 10^{-2} \text{ J} = 0.2 \text{ J}\)