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Two parallel plate capacitors differ only in the spacing between their (very thin) plates; one, AB, has a spacing of 5 mm and a capacitance of 20 pF, the other, CD, has a spacing of 2 mm. Plates A and C carry charges of +1 nC, whilst B and D each carry −1 nC. What are the potential differences {{V}_{AB}} and {{V}_{CD}} after the capacitor CD is slid centrally between and parallel to the plates of AB without touching them? Would it make any difference if CD were not centrally placed between A and B?

Step-by-step

The net charges on the plates cannot change, but the charges on the plates on either side of any of the spaces must be equal and opposite. Consequently, the charges on C and D must be −1 nC and +1 nC, respectively, on their outside surfaces and +2 nC and −2 nC, respectively, on their inside surfaces. The capacitance of any pair of plates is inversely proportional to their separation, with 5 mm corresponding to 20 pF.
Thus if AC is x mm and DB is (3 − x) mm, the capacitances of the three successive capacitors are 100/x, 50 and 100/(3 − x) pF. The voltage {{V}_{CD}} is therefore 40 V, and {{V}_{AB}} is 10x + 40 + 10(3 − x) = 70 V, independent of the value of x.

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