A parallel-plate capacitor is made by inserting a sheet of dielectric material between two plates - AQA - A-Level Physics - Question 19 - 2019 - Paper 2

To determine which combination of relative permittivity and thickness yields the greatest capacitance, we evaluate based on the formula for capacitance:

C = rac{εA}{d}

where:

• $C$ is the capacitance,
• $ε$ is the permittivity of the dielectric material (which is relative permittivity $ε_r$ multiplied by the permittivity of free space $ε_0$),
• $A$ is the area of the plates,
• $d$ is the thickness of the dielectric material.

Given the relative permittivities and thicknesses in the table:

• A: $ε_r = 2$, $d = 0.40 \text{ mm}$
• B: $ε_r = 3$, $d = 0.90 \text{ mm}$
• C: $ε_r = 4$, $d = 1.0 \text{ mm}$
• D: $ε_r = 6$, $d = 1.6 \text{ mm}$

Calculating the effective capacity for each:

For option A: $ext{Effective } C_A = \frac{(2ε_0)A}{0.40 \text{ mm}}$

For option B: $ext{Effective } C_B = \frac{(3ε_0)A}{0.90 \text{ mm}}$

For option C: $ext{Effective } C_C = \frac{(4ε_0)A}{1.0 \text{ mm}}$

For option D: $ext{Effective } C_D = \frac{(6ε_0)A}{1.6 \text{ mm}}$

Now we need to compare these ratios. Given that higher relative permittivity and lower thickness maximize capacitance, we find:

• Comparing A and D: A has the lowest thickness, D has the highest relative permittivity, but A's capacitive effect is greater due to the smallest thickness.

Thus, the best combination is:

Answer: Option D (relative permittivity = 6 and thickness = 1.6 mm) for maximal capacitance.