Capacitors

JEE Main level, single correct, medium

13 of 180

Question 13 of 180

Two identical thin metal plates has charge q1{q}_{1} and q2{q}_{2} respectively such that q1>q2{q}_{1}\gt {q}_{2}. The plates were brought close to each other to form a parallel plate capacitor of capacitance CC. The potential difference between them is :

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Answer: (C) (q1−q2)2C\dfrac{\left({q}_{1}-{q}_{2}\right)}{2C}

On bringing the charged metal plates closer, electric field E⃗\vec{E} in the intervening space is E⃗=E1⃗+E2⃗\vec{E}=\vec{{E}_{1}}+\vec{{E}_{2}}
Where
Intensity of field due plate charged by q1{q}_{1} is ∣E1⃗∣=σ12ϵ0=q12ϵ0A\left|\vec{{E}_{1}}\right|=\dfrac{{\sigma }_{1}}{2{\epsilon }_{0}}=\dfrac{{q}_{1}}{2{\epsilon }_{0}A}(directed rightwards)
And Intensity of field due to plate charged by q2{q}_{2} is
∣E2⃗∣=σ22ϵ0=q22ϵ0A\left|\vec{{E}_{2}}\right|=\dfrac{{\sigma }_{2}}{2{\epsilon }_{0}}=\dfrac{{q}_{2}}{2{\epsilon }_{0}A} (directed leftwards)
So, Net field is given by
E⃗=E1⃗+E2⃗⇒E=E1−E2\vec{E}=\vec{{E}_{1}}+\vec{{E}_{2}}\Rightarrow E={E}_{1}-{E}_{2}
∴E=q12ϵ0A−q22ϵ0A=(q1−q2)2ϵ0A...(1)∴E=\dfrac{{q}_{1}}{2{\epsilon }_{0}A}-\dfrac{{q}_{2}}{2{\epsilon }_{0}A}=\dfrac{\left({q}_{1}-{q}_{2}\right)}{2{\epsilon }_{0}A}...\left(1\right)
For parallel plate capacitor C=ϵ0Ad...(2)C=\dfrac{{\epsilon }_{0}A}{d}...\left(2\right)
From above two equations
E=(q1−q2)2Cd...(3)E=\dfrac{\left({q}_{1}-{q}_{2}\right)}{2C}d...\left(3\right)
∵∵Relation between intensity of field (E)\left(E\right) and potential difference (V)\left(V\right) between the plates is
E=Vd...(4)E=\dfrac{V}{d}...\left(4\right)
From equation (3)\left(3\right) and (4)\left(4\right)
V=(q1−q2)2CV=\dfrac{\left({q}_{1}-{q}_{2}\right)}{2C}

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