Two identical thin metal plates has charge q 1 and q 2 respectively such that q 1 q 2 . The plates were brought close to each other to form a parallel plate capacitor of capacitance C . The potentia

Question

Two identical thin metal plates has charge q 1 and q 2 respectively such that q 1 > q 2 . The plates were brought close to each other to form a parallel plate capacitor of capacitance C . The potential difference between them is :

TestHub · Accepted Answer

On bringing the charged metal plates closer, electric field $\vec{E}$ in the intervening space is $\vec{E} = \vec{E_{1}} + \vec{E_{2}}$
Where
Intensity of field due plate charged by $q_{1}$ is $|\vec{E_{1}}| = \frac{σ_{1}}{2 ε_{0}} = \frac{q_{1}}{2 ε_{0} A}$ (directed rightwards)
And Intensity of field due to plate charged by $q_{2}$ is
$|\vec{E_{2}}| = \frac{σ_{2}}{2 ε_{0}} = \frac{q_{2}}{2 ε_{0} A}$ (directed leftwards)
So, Net field is given by
$\vec{E} = \vec{E_{1}} + \vec{E_{2}} \Rightarrow E = E_{1} - E_{2}$
$∴ E = \frac{q_{1}}{2 ε_{0} A} - \frac{q_{2}}{2 ε_{0} A} = \frac{(q_{1} - q_{2})}{2 ε_{0} A} . . . (1)$
For parallel plate capacitor $C = \frac{ε_{0} A}{d} . . . (2)$
From above two equations
$E = \frac{(q_{1} - q_{2})}{2 C} d . . . (3)$
$∵$ Relation between intensity of field $(E)$ and potential difference $(V)$ between the plates is
$E = \frac{V}{d} . . . (4)$
From equation $(3)$ and $(4)$
$V = \frac{(q_{1} - q_{2})}{2 C}$

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