AP Intermediate 2nd Year Electric Charges and Fields — practice questions — page 3

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 $C = \frac{\epsilon_0 A}{d}$.

1. U = ½ C V². 2. U = 0.5 × 5 × 10⁻⁶ × (200)² = 0.5 × 5 × 10⁻⁶ × 4 × 10⁴. 3. U = 10⁻¹ = 0.1 J. _Source: syllabus Physics Part I, Ch 2 §2.15_

1. Potential is defined up to an additive constant; choosing V(∞) = 0 fixes the constant. 2. With this choice, V due to a +Q point charge is positive everywhere finite. 3. The other options have no physical basis. _Source: syllabus Physics Part I, Ch 2 §2.3_