A parallel-plate capacitor has square plates of length l separated by distance d and is filled with a dielectric - AQA - A-Level Physics - Question 24 - 2018 - Paper 2

For the second capacitor, where the plates have length 2l and are separated by distance 2d, the capacitance is:

C_2 = rac{ ext{ε}_0 imes A'}{d'} where:

• A' = (2l)^2 = 4l^2,
• d' = 2d,
• ε₀ represents the permittivity of free space (or air).

Thus, for the second capacitor: C_2 = rac{ ext{ε}_0 imes 4l^2}{2d} = rac{2 ext{ε}_0 imes l^2}{d}

Since both capacitors have the same capacitance, $C_1 = C_2$ we can write:

rac{ ext{ε} imes l^2}{d} = rac{2 ext{ε}_0 imes l^2}{d}

This simplifies to: $ext{ε} = 2 ext{ε}_0$

The relative permittivity (κ) is defined as: κ = rac{ ext{ε}}{ ext{ε}_0}

Thus: $κ = 2$

Therefore, the relative permittivity of the dielectric in the first capacitor is: C. 2