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: Calculate the extra charge.

The change in charge (\(\Delta q\)) due to the introduction of the dielectric is given by:

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

where K is the dielectric constant, C is the capacitance, and V is 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 change in electrostatic energy.

The change in electrostatic energy (\(\Delta U\)) is given by:

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

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

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

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

Answer 2

Step 1 — Initial charge on the capacitor:
The capacitance without dielectric is $C_1 = 40\,\mu F$.
The voltage is $V = 100\,V$.
So, initial charge:
$$ Q_1 = C_1 V = 40 \times 10^{-6} \times 100 = 4.0 \times 10^{-3}\,C = 4\,mC $$

Step 2 — New capacitance after inserting dielectric:
$$ C_2 = K C_1 = 2 \times 40 = 80\,\mu F $$

Step 3 — New charge on the capacitor:
Since the voltage source is still connected (constant $V$):
$$ Q_2 = C_2 V = 80 \times 10^{-6} \times 100 = 8.0 \times 10^{-3}\,C = 8\,mC $$

Step 4 — Extra charge supplied by the battery:
$$ \Delta Q = Q_2 - Q_1 = 8 - 4 = 4\,mC $$

Step 5 — Change in electrostatic energy:
For constant potential, energy stored in a capacitor is $U = \dfrac{1}{2} C V^2$.

Initial energy: $$ U_1 = \frac{1}{2} \times 40 \times 10^{-6} \times (100)^2 = 0.2\,J $$
Final energy: $$ U_2 = \frac{1}{2} \times 80 \times 10^{-6} \times (100)^2 = 0.4\,J $$
Change in energy: $$ \Delta U = U_2 - U_1 = 0.4 - 0.2 = 0.2\,J $$

Step 6 — Final Results:
Extra charge $= 4\,mC$
Increase in energy $= 0.2\,J$

✅ Correct Option: 3