1. Initial capacitor data

Given:

• Initial capacitance: $C = 40\,\mu\text{F} = 40 \times 10^{-6}\,\text{F}$
• Battery voltage: $V = 100\,\text{V}$
• Dielectric constant: $K = 2$

Since the capacitor remains connected to the power supply, the voltage remains constant.

2. Initial charge on the capacitor

Using $Q_i = CV$ we get $Q_i = 40 \times 10^{-6} \times 100 = 4 \times 10^{-3}\,\text{C} = 4\,\text{mC}$

3. New capacitance after inserting dielectric

For a dielectric fully filling the space, $C' = KC = 2 \times 40\,\mu\text{F} = 80\,\mu\text{F}$

4. Final charge on the capacitor

Since voltage is still $100\,\text{V}$, $Q_f = C'V = 80 \times 10^{-6} \times 100 = 8 \times 10^{-3}\,\text{C} = 8\,\text{mC}$

So the extra charge is $\Delta Q = Q_f - Q_i = 8\,\text{mC} - 4\,\text{mC} = 4\,\text{mC}$

5. Initial electrostatic energy

For constant voltage, $U_i = \frac{1}{2}CV^2$ Thus, $U_i = \frac{1}{2}(40 \times 10^{-6})(100)^2$ $U_i = \frac{1}{2}(40 \times 10^{-6})(10000)$ $U_i = 0.2\,\text{J}$

6. Final electrostatic energy

$U_f = \frac{1}{2}C'V^2 = \frac{1}{2}(80 \times 10^{-6})(100)^2$ $U_f = 0.4\,\text{J}$

Hence the change in electrostatic energy is $\Delta U = U_f - U_i = 0.4 - 0.2 = 0.2\,\text{J}$

7. Match with options

• Extra charge $= 4\,\text{mC}$
• Change in electrostatic energy $= 0.2\,\text{J}$

So the correct option is:

$\boxed{\text{B}}$

8. Comparison with stored answer

Stored correct answer: B

My derived answer also gives B, so they agree.