A parallel plate capacitor of capacitance \( 12.5 \, \text{pF} \) is charged by a battery connected between its plates to a potential difference of \( 12.0 \, \text{V} \). The battery is now disconnected and a dielectric slab (\( \varepsilon_r = 6 \)) is inserted between the plates. The change in its potential energy after inserting the dielectric slab is ______ \( \times 10^{-12} \, \text{J} \).
Correct Answer: 750
Approach Solution - 1
1. Initial Capacitance and Charge on Capacitor:
The initial capacitance \( C_0 = 12.5 \, \text{pF} \), and the initial charge on the capacitor \( Q = C_0V \).
2. Capacitance with Dielectric Inserted:
After inserting a dielectric with dielectric constant \( \epsilon_r = 6 \), the new capacitance becomes:
\[ C_f = \epsilon_r C_0. \]
3. Change in Potential Energy:
The change in potential energy of the capacitor is given by:
\[ \Delta U = U_i - U_f = \frac{Q^2}{2C_i} - \frac{Q^2}{2C_f}. \] Substituting \( Q = C_0V \) and simplifying:
\[ \Delta U = \frac{(C_0V)^2}{2C_0} \left[ 1 - \frac{1}{\epsilon_r} \right] = \frac{1}{2} C_0V^2 \left[ 1 - \frac{1}{6} \right]. \]
4. Calculation:
Substitute \( C_0 = 12.5 \, \text{pF}, \, V = 12 \, \text{V}, \, \text{and} \, \epsilon_r = 6 \):
\[ \Delta U = \frac{1}{2} \times 12.5 \times 10^{-12} \times (12)^2 \times \frac{5}{6}. \] Simplifying further:
\[ \Delta U = 750 \, \text{pJ} = 750 \times 10^{-12} \, \text{J}. \]
Answer: \( 750 \times 10^{-12} \, \text{J} \)
Approach Solution -2
Capacitance of the capacitor before inserting dielectric, \( C_1 = 12.5 \, \text{pF} = 12.5 \times 10^{-12} \, \text{F} \)
Potential difference, \( V_1 = 12.0 \, \text{V} \)
Dielectric constant, \( \varepsilon_r = 6 \)
Battery is disconnected (which means the charge on the capacitor remains constant).
Step 2: Recall expressions for potential energy of a capacitor
The energy stored in a charged capacitor is given by:
\[ U = \frac{1}{2} C V^2 = \frac{Q^2}{2C} \] Since the battery is disconnected, \( Q \) remains constant.
Therefore, the change in potential energy is due to the change in capacitance after inserting the dielectric slab.
Step 3: Calculate the charge on the capacitor before inserting the dielectric
\[ Q = C_1 V_1 = 12.5 \times 10^{-12} \times 12 = 150 \times 10^{-12} \, \text{C} \] \[ Q = 1.5 \times 10^{-10} \, \text{C} \]
Step 4: Find the new capacitance after inserting the dielectric
When the dielectric slab of dielectric constant \( \varepsilon_r \) is inserted, the new capacitance becomes:
\[ C_2 = \varepsilon_r C_1 = 6 \times 12.5 \times 10^{-12} = 75 \times 10^{-12} \, \text{F} \]
Step 5: Calculate the potential energies before and after insertion
Before insertion:
\[ U_1 = \frac{Q^2}{2C_1} \] \[ U_1 = \frac{(1.5 \times 10^{-10})^2}{2 \times 12.5 \times 10^{-12}} = \frac{2.25 \times 10^{-20}}{25 \times 10^{-12}} = 0.9 \times 10^{-9} = 900 \times 10^{-12} \, \text{J} \]
After insertion:
\[ U_2 = \frac{Q^2}{2C_2} \] \[ U_2 = \frac{(1.5 \times 10^{-10})^2}{2 \times 75 \times 10^{-12}} = \frac{2.25 \times 10^{-20}}{150 \times 10^{-12}} = 0.15 \times 10^{-9} = 150 \times 10^{-12} \, \text{J} \]
Step 6: Calculate the change in potential energy
\[ \Delta U = U_2 - U_1 = 150 \times 10^{-12} - 900 \times 10^{-12} = -750 \times 10^{-12} \, \text{J} \] The negative sign indicates that the potential energy decreases after the dielectric is inserted.
Step 7: Interpretation
Since the battery is disconnected, the charge stays the same, and inserting a dielectric increases capacitance. As a result, the stored energy decreases because the same charge now has a lower potential difference across the plates.
Final Answer (magnitude of change):
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