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 : (1) 2 mC and 0.2 J (2) 8 mC and 2.0 J (3) 4 mC and 0.2 J (4) 2 mC and 0.4 J

Key ideas

1. Capacitance with a dielectric
   A parallel-plate capacitor’s capacitance multiplies by the dielectric constant $K$: $C_{new} = K \times C_{old}$

2. Battery remains connected
   Because the 100 V battery stays attached, the voltage stays the same before and after insertion. What changes are capacitance and therefore charge and stored energy.

3. Charge stored on a capacitor
   $Q = C V$

4. Energy stored in a capacitor
   $U = \frac{1}{2} C V^2$

Logical steps a student should follow

1. Start with the given capacitance $C_0 = 40\,\mu F$ and voltage $V = 100\,V$.
2. Compute the initial charge $Q_0$ using $Q = CV$.
3. Compute the initial energy $U_0$ using $U = \tfrac12 C V^2$.
4. Insert the dielectric ($K = 2$) to get the new capacitance $C_1 = 2C_0$.
5. With the same voltage, find the new charge $Q_1$ and the new energy $U_1$.
6. Find the extra charge $\Delta Q = Q_1 - Q_0$ and the change in energy $\Delta U = U_1 - U_0$.
7. Compare results with the options provided.

Imagine this!

1. You have a bucket that can hold water.
2. At first it holds 40 cups of water when you pour it up to a certain height (this is like a 40 µF capacitor holding charge at 100 V).
3. Now you slide in a magic sponge (the dielectric, $K = 2$) that makes the bucket able to hold twice as much water while the tap (battery) still pours at the same pressure (voltage).
4. Because the bucket can now hold more, extra water flows in (extra charge).
5. More water up high means more stored energy (energy in a capacitor).

We want to know:

• How much extra water (charge) flowed in?
• How much more energy got stored?

Step-by-step solution

1. Initial data:
   $C_0 = 40\,\mu F = 40 \times 10^{-6}\,F$
   $V = 100\,V$

2. Initial charge:
   $Q_0 = C_0 V = (40 \times 10^{-6})(100) = 4 \times 10^{-3}\,C = 4\,mC$

3. Initial energy:

   $$U_0 = \frac{1}{2} C_0 V^2 = \frac{1}{2}(40 \times 10^{-6})(100)^2 \\ = \frac{1}{2}(40 \times 10^{-6})(10^4) = \frac{1}{2}(40)(10^{-6})(10^4) \\ = \frac{1}{2}(40)(10^{-2}) = 0.2\,J$$

4. Insert dielectric ($K = 2$) → new capacitance:
   $C_1 = K C_0 = 2 \times 40\,\mu F = 80\,\mu F$

5. New charge (battery still 100 V):
   $Q_1 = C_1 V = (80 \times 10^{-6})(100) = 8 \times 10^{-3}\,C = 8\,mC$

6. Extra charge:
   $\Delta Q = Q_1 - Q_0 = 8\,mC - 4\,mC = 4\,mC$

7. New energy:

   $$U_1 = \frac{1}{2} C_1 V^2 = \frac{1}{2}(80 \times 10^{-6})(100)^2 \\ = \frac{1}{2}(80)(10^{-2}) = 0.4\,J$$

8. Change in energy:
   $\Delta U = U_1 - U_0 = 0.4\,J - 0.2\,J = 0.2\,J$

9. Match with options:
   Option (3) gives 4 mC extra charge and 0.2 J energy change.

Answer: Option (3) 4 mC and 0.2 J.