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Question

Two parallel plate capacitors (AB & CD) differ only in the spacing between their plates(very thin).
Capacitor AB has a spacing of 5 mm and a capacitance of 20 pF, the capacitor CD, has a spacing of 2 mm.
Plates A and C carry charges of +1nC, while B and D each carry 1nC.
If the capacitor CD is slid centrally between and parallel to the plates of AB without touching them (as shown below), what is the potential difference VAB ?


A
70 V
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B
50 V
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C
40 V
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D
None of these
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Solution

The correct option is A 70 V
Charge re-distribution on the plates is as shown below:


VAC=QCAC=QdCAϵ0A
VDB=QCDB=Q[dDBϵ0A]
VAC+VDB=QdCA+dDBϵ0A
VAC+VDB=Q(3)ϵ0A=Q5ϵ0A×35=1 nC20 pF×35
VAC+VDB=30 V ...(1)

VCD=QCCD=2 nC20 pF×52=40 V ...(2)
On adding (1) and (2), we get VAB=30 V+40 V=70 V

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