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Question

A parallel-plate capacitor has plates of area 200 cm2 and separation between the plates 1 mm. The potential difference developed if a charge of 1 nC(i.e.1×109C) is given to the capacitor is something,then if the plate separation is now increased to 2 mm, what will be the new potential difference?

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Solution

We know that
QC=V .......(i)
And also C=ε0Ad
Putting this in eq(i)
V=1×109×1×103200×104×8.85×1012=5.65 V
If plate separation(d) is made 2 mm then potential difference (V) is given by
V=QC=1×109×2×103200×104×8.85×1012=11.3 V

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