Karnataka PUC II Electric Charges and Fields — practice questions — page 3

_Source: syllabus Physics Electrostatic Potential and Capacitance, electrostatic potential_

To derive the capacitance of a parallel-plate capacitor, we follow these steps: 1. **Define Capacitance**: Capacitance $C$ is the ratio of the magnitude of the charge $Q$ on either conductor to the magnitude of the potential difference $V$ between them: $$C = \frac{Q}{V}$$ 2. **Find the Electric Field**: For a parallel-plate capacitor with area $A$ and charge $Q$, the surface charge density is $\sigma = Q/A$. The electric field $E$ between the plates is uniform and given by: $$E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}$$ 3. **Relate Field to Potential**: The potential difference $V$ between plates separated by distance $d$ is: $$V = E \cdot d = \frac{Q \cdot d}{\epsilon_0 A}$$ 4. **Calculate Capacitance**: Substitute $V$ back into the capacitance formula: $$C = \frac{Q}{V} = \frac{Q}{\left(\frac{Q \cdot d}{\epsilon_0 A}\right)} = \frac{\epsilon_0 A}{d}$$ Thus, the capacitance of a parallel-plate capacitor is $C = \frac{\epsilon_0 A}{d}$.