A parallel-plate capacitor of capacitance $$40\mu$$ 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

A parallel-plate capacitor of capacitance $$40\mu F$$ is connected to a 100 V supply. A dielectric of K=2 is inserted while the supply remains connected.

Initial capacitance: $$C_0 = 40 \mu F$$

$$Q_0 = C_0 V = 40 \times 10^{-6} \times 100 = 4 \times 10^{-3} \text{ C} = 4 \text{ mC}$$

$$U_0 = \frac{1}{2}C_0 V^2 = \frac{1}{2} \times 40 \times 10^{-6} \times (100)^2 = 0.2 \text{ J}$$

New capacitance: $$C = KC_0 = 2 \times 40 = 80 \mu F$$

$$Q = CV = 80 \times 10^{-6} \times 100 = 8 \times 10^{-3} \text{ C} = 8 \text{ mC}$$

$$U = \frac{1}{2}CV^2 = \frac{1}{2} \times 80 \times 10^{-6} \times (100)^2 = 0.4 \text{ J}$$

Extra charge: $$\Delta Q = Q - Q_0 = 8 - 4 = 4 \text{ mC}$$

Change in energy: $$\Delta U = U - U_0 = 0.4 - 0.2 = 0.2 \text{ J}$$

The correct answer is Option 1: 4 mC and 0.2 J.