A parallel-plate capacitor of capacitance $40\mathrm{\mu F}$ 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 -

Step 1: Capacitance After Inserting the Dielectric

When a dielectric of constant $K$ is inserted, the new capacitance increases as:

$$C_2 = K C_1$$

 $$C_2 = 2 \times 40 \mu F = 80 \mu F$$ 

Step 2: Extra Charge Stored

Since the capacitor is connected to the same voltage source, the voltage remains constant ($V = 100V$). The charge stored in a capacitor is given by:

$$Q = CV$$

Initial charge (before dielectric):

$$Q_1 = C_1 V = (40 \times 10^{-6}) \times 100 = 4 \times 10^{-3} C = 4 mC$$

Final charge (after dielectric):

$$Q_2 = C_2 V = (80 \times 10^{-6}) \times 100 = 8 \times 10^{-3} C = 8 mC$$

Extra charge added:

$$\Delta Q = Q_2 - Q_1 = 8 mC - 4 mC = 4 mC$$

Step 3: Change in Electrostatic Energy

Electrostatic energy stored in a capacitor is given by:

$$U = \frac{1}{2} C V^2$$

Initial energy (before dielectric):

$$U_1 = \frac{1}{2} \times 40 \times 10^{-6} \times (100)^2$$ 

$$U_1 = \frac{1}{2} \times 40 \times 10^{-6} \times 10000$$

$$U_1 = 0.2 \text{ J}$$

Final energy (after dielectric):

$$U_2 = \frac{1}{2} \times 80 \times 10^{-6} \times (100)^2$$

$$U_2 = \frac{1}{2} \times 80 \times 10^{-6} \times 10000$$

$$U_2 = 0.4 \text{ J}$$

Change in electrostatic energy:

$$\Delta U = U_2 - U_1 = 0.4 J - 0.2 J = 0.2 J$$