The distance between the plates of a plane capacitor is d and the area of each plate is A. As shown in the figure, both plates of the capacitor are earthed and a small body carrying charge Q is placed between them, at a distance x from one plate. What charge will accumulate on each plate?

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Accepted Answer

The charge between the capacitor plates could be notionally divided into two parts, which are then moved away from each other in a direction parallel to the plates. The (induced) charges on the capacitor plates would also move but their totals on each plate would remain unaltered (using the principle of superposition). Continuing in the same manner, the charge Q could be further subdivided until it was ‘spread’ uniformly on a plane of the same size as, and parallel to, the capacitor plates.
We can now consider two plane capacitors connected in parallel, the distances between their plates being x and d − x. Since one plate of each of the capacitors is earthed, and the other is common, the voltages between the pairs of plates are identical. Their electric field strengths are therefore inversely proportional to their plate separations, i.e. [latex] {{E}_{1}}/{{E}_{2}} [/latex] = (d − x)/x. The electric flux emanating from the charge Q is divided in the same proportion (Gauss’s law), and the ratio of the charges on the earthed plates is the same as well. Since a total charge of −Q accumulates on the plates, respective charges of [latex]{{Q}_{1}}=-Q{\frac {d-x} {d}} [/latex] and [latex] {{Q}_{2}}=-Q{\frac {x} {d}} [/latex] accumulate on the two separate plates.

Question : The distance between the plates of a plane capacitor is d an...

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