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Question

(C) When two large conducting plates with charges $q_{1}$ and $q_{2}$ are placed parallel to each other,the charge on the inner surfaces is given by $

Vedclass · Accepted Answer

$\frac{(q_{1}-q_{2})}{2C}$

Vedclass · Answer

(C) When two large conducting plates with charges $q_{1}$ and $q_{2}$ are placed parallel to each other,the charge on the inner surfaces is given by $q_{inner} = \frac{q_{1}-q_{2}}{2}$.The electric field $E$ between the plates is due to these inner charges: $E = \frac{\sigma}{\varepsilon_{0}} = \frac{q_{inner}}{A\varepsilon_{0}} = \frac{q_{1}-q_{2}}{2A\varepsilon_{0}}$.The potential difference $V$ between the plates is given by $V = E \cdot d$,where $d$ is the separation between the plates.Since the capacitance of a parallel plate capacitor is $C = \frac{A\varepsilon_{0}}{d}$,we can write $V = \frac{q_{1}-q_{2}}{2A\varepsilon_{0}} \cdot d = \frac{q_{1}-q_{2}}{2(A\varepsilon_{0}/d)}$.Substituting $C$,we get $V = \frac{q_{1}-q_{2}}{2C}$.

Two identical thin metal plates have charges $q_{1}$ and $q_{2}$ respectively,such that $q_{1} > q_{2}$. The plates are brought close to each other to form a parallel plate capacitor of capacitance $C$. The potential difference between them is:

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