The distance between the two plates of a parallel plate capacitor is doubled, and the area of each plate is ha

Question

The distance between the two plates of a parallel plate capacitor is doubled, and the area of each plate is halved. If C is its initial capacitance, its final capacitance is equal to:

Accepted Answer

1. **Formula:** The capacitance of a parallel plate capacitor is given by the formula $C = \frac{\varepsilon_0 A}{d}$, where $A$ is the area of the plates and $d$ is the separation distance.
2. **Initial State:** $C_{initial} = C = \frac{\varepsilon_0 A}{d}$.
3. **New Parameters:** 
   - New distance $d' = 2d$.
   - New area $A' = \frac{A}{2}$.
4. **Final Capacitance:** Substitute the new values into the formula:
   $C_{final} = \frac{\varepsilon_0 A'}{d'} = \frac{\varepsilon_0 (A/2)}{2d} = \frac{\varepsilon_0 A}{4d} = \frac{1}{4} \left( \frac{\varepsilon_0 A}{d} \right)$.
5. **Conclusion:** $C_{final} = \frac{C}{4}$.

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