A capacitor of capacitance 63 pF is made from two parallel metal plates separated by an air gap - AQA - A-Level Physics - Question 1 - 2021 - Paper 2

Polar molecules within mica have a positive and negative side, which align themselves with an external electric field. When the mica is inserted between the capacitor plates, these molecules orient themselves such that they oppose the electric field. This opposition reduces the effective electric field between the plates, allowing the capacitor to store more charge for the same voltage. Consequently, the capacitance increases, as it is defined by the relationship: $C = \frac{Q}{V}$ where Q is the charge stored and V is the voltage across the plates.

The initial energy stored in the capacitor (without mica) can be calculated using the formula: $E_i = \frac{1}{2} C V^2$ Given that the capacitance C = 63 pF, and the charge Q = 7.6 × 10⁻¹⁰ C, we can find V using the relation: $V = \frac{Q}{C} = \frac{7.6 \times 10^{-10}}{63 \times 10^{-12}} \approx 12.06 \text{ volts}$ Computing the initial energy: $E_i = \frac{1}{2} (63 \times 10^{-12}) (12.06^2) \approx 4.6 \times 10^{-10} ext{ joules}$ When the mica is fully inserted, the capacitance becomes: $C_f = 6 (63 pF) = 378 pF$ The new voltage when fully inserted is given by $V_f = \frac{Q}{C_f} = \frac{7.6 \times 10^{-10}}{378 \times 10^{-12}} \approx 2.01 \text{ volts}$ Thus, the final energy becomes: $E_f = \frac{1}{2} (378 \times 10^{-12}) (2.01^2) \approx 0.76 \times 10^{-10} ext{ joules}$ Finally, the difference in energy is: $\Delta E = E_i - E_f \approx (4.6 \times 10^{-10} - 0.76 \times 10^{-10}) \approx 3.84 \times 10^{-10} ext{ joules}$

The graph should be a linear increase, starting from 0 when θ = 0° where the plates fully overlap, and reaching a maximum value at θ = 360°, where the plates are entirely separated. The x-axis will represent the angle θ in degrees, and the y-axis will represent the capacitance C.

1. Using Half the Area: The capacitance formula is given by: $C = \frac{\varepsilon A}{d}$ If the diameter of the capacitor plates is reduced by half, the area A is reduced by a factor of 1/4. To maintain the same capacitance, the distance d between the plates could be halved, yielding an effective increase in capacitance.

2. Increasing Dielectric Constant: Another approach would be to replace air with a dielectric material with a higher dielectric constant than air (which is approximately 1), effectively allowing for the same capacitance value with the reduced plate area.