Question:

A parallel plate capacitor consists of two square plates each with side length 's' and separated by a distance 'd'. The space between the plates is filled with a dielectric material of dielectric constant 'k'. The capacitor is connected to a battery with potential difference 'V' across it.

a) Calculate the capacitance of the capacitor in terms of 's', 'd', and 'k'.

b) If the distance between the plates is doubled while the dielectric and plate dimensions remain the same, how does the capacitance change? Explain your answer.

Answer:

a) The formula to calculate the capacitance of a parallel plate capacitor is given by:

Where:

• C: capacitance [Farads (F)]
• ε₀: permittivity of free space = 8.85 x 10⁻¹² F/m
• A: area of each plate = s² [m²]
• k: dielectric constant (relative permittivity)

Given that the side length of each square plate is 's' and the distance between the plates is 'd', the area of each plate is A = s². Substituting these values into the capacitance formula, we get:

b) When the distance between the plates is doubled while the dielectric and plate dimensions remain the same, the capacitance of the capacitor also doubles. This is because the capacitance is inversely proportional to the distance between the plates (C ∝ 1/d). When the distance 'd' doubles, the denominator in the capacitance formula also doubles, leading to a doubling of the overall capacitance.