Question

An isolated parallel-plate capacitor of area \({A_1}\) with an air gap of length \({s_1}\) is charged up to a potential difference\(\Delta {V_1}\). A second parallel-plate capacitor, initially uncharged, has an area \({A_2}\) and a gap of length \({s_2}\)filled with plastic whose dielectric constant is\(K\). You connect a wire from the positive plate of the first capacitor to one of the second capacitor, and you connect another wire from the negative plate of the first capacitor to the other plate of the second capacitor. What is the final potential difference across the first capacitor?

Short answer

\[\Delta V = \frac{{{A_1}{s_2}\Delta {V_1}}}{{{A_1}{s_2} + K{A_2}{s_1}}}\]

Step-by-step solution

Given data

Capacitor1: Area=\({A_1}\), Gap=\({s_1}\), Potential diff. = \(\Delta {V_1}\)

Capacitor2: Area= A₂, Gap= s₂, Dielectric const. = K

Concept/ Formula used

There is a potential difference between a capacitor's plates when it is completely charged, and the larger the area of the plates and/or the closer together they are, the more charge the capacitor can keep and the greater the capacitance.

\(C = \frac{Q}{{\Delta V}} = \frac{{{\varepsilon _0}A}}{s}\)

Where,\(C\)is capacitance,\(A\)is area of plate,\(s\)is gap between the plate and\(\Delta V\)is potential difference.

Derivation for final potential difference

Initial charge on first plate = \({q_1} = \frac{{\varepsilon {A_1}}}{{{s_1}}}\Delta {V_1}\)

Initial charge on second plate = \({q_2} = 0\) (initially uncharged)

Let final potential difference = \(\Delta V\)

Final charge \(Q = {q_1} + {q_2}\)

.\[\begin{aligned}\left( {\frac{{\varepsilon {A_1}}}{{{s_1}}} + \frac{{K\varepsilon {A_2}}}{{{s_2}}}} \right)\Delta V &= \frac{{\varepsilon {A_1}}}{{{s_1}}}\Delta {V_1} + 0\\\left( {\frac{{{A_1}}}{{{s_1}}} + \frac{{K{A_2}}}{{{s_2}}}} \right)\Delta V &= \frac{{{A_1}}}{{{s_1}}}\Delta {V_1}\\\Delta V &= \frac{{{A_1}{s_2}\Delta {V_1}}}{{{A_1}{s_2} + K{A_2}{s_1}}}\end{aligned}\].