Two identical metal plates are given charges | vedclass

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(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_{

Vedclass · Accepted Answer

$\frac{q_1 - q_2}{2C}$

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(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 $q_{inner} = \frac{q_1 - q_2}{2}$.This charge $q_{inner}$ is responsible for the electric field between the plates.The electric field $E$ between the plates is given by $E = \frac{q_{inner}}{A \epsilon_0} = \frac{q_1 - q_2}{2A \epsilon_0}$.The potential difference $V$ between the plates is $V = E \cdot d$, where $d$ is the distance between the plates.Since the capacitance of a parallel plate capacitor is $C = \frac{A \epsilon_0}{d}$, we have $\frac{1}{C} = \frac{d}{A \epsilon_0}$.Substituting this into the expression for $V$, we get $V = \frac{q_1 - q_2}{2A \epsilon_0} \cdot d = \frac{q_1 - q_2}{2} \cdot \frac{d}{A \epsilon_0} = \frac{q_1 - q_2}{2C}$.

Two identical metal plates are given charges $q_1$ and $q_2$ $(q_2 < q_1)$ respectively. They are brought close together to form a parallel plate capacitor with capacitance $C$. The potential difference $V$ between the plates is

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