29. A parallel plate capacitor of capacitance \( 40\,\mu\text{F} \) is connected to a \( 100\,\text{V} \) power supply.
Now the intermediate space between the plates is completely filled with a dielectric material of dielectric constant \( K = 2 \).
Due to the introduction of the dielectric material, the extra charge and the change in the electrostatic energy stored in the capacitor are, respectively:
- 2 mC and 0.4 J
- 2 mC and 0.2 J
- 4 mC and 0.2 J
- 8 mC and 2.0 J
Answer: 3. 4 mC and 0.2 J
Solution: Since the capacitor remains connected to the battery, the potential difference remains constant: \( V = 100\,\text{V} \).
Initial capacitance: \( C_0 = 40\,\mu\text{F} = 40 \times 10^{-6}\,\text{F} \)
Initial charge: \( Q_0 = C_0 V = 40 \times 10^{-6} \times 100 = 4 \times 10^{-3}\,\text{C} = 4\,\text{mC} \)
After inserting dielectric (\( K = 2 \)):
New capacitance: \( C = K C_0 = 2 \times 40 = 80\,\mu\text{F} \)
New charge: \( Q = C V = 80 \times 10^{-6} \times 100 = 8 \times 10^{-3}\,\text{C} = 8\,\text{mC} \)
Extra charge supplied by battery:
\( \Delta Q = Q – Q_0 = 8\,\text{mC} – 4\,\text{mC} = 4\,\text{mC} \)
Initial energy stored:
\( U_0 = \frac{1}{2} C_0 V^2 = \frac{1}{2} \times 40 \times 10^{-6} \times (100)^2 = 0.2\,\text{J} \)
Final energy stored:
\( U = \frac{1}{2} C V^2 = \frac{1}{2} \times 80 \times 10^{-6} \times (100)^2 = 0.4\,\text{J} \)
Change in electrostatic energy stored:
\( \Delta U = U – U_0 = 0.4\,\text{J} – 0.2\,\text{J} = +0.2\,\text{J} \)
Therefore, extra charge = 4 mC and change in energy = +0.2 J