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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.1×10−9C) 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=1×10−9×1×10−3200×10−4×8.85×10−12=5.65 V
If plate separation(d) is made 2 mm then potential difference (V′) is given by
V′=QC=1×10−9×2×10−3200×10−4×8.85×10−12=11.3 V
QC=V .......(i)
And also C=ε0Ad
Putting this in eq(i)
V=1×10−9×1×10−3200×10−4×8.85×10−12=5.65 V
If plate separation(d) is made 2 mm then potential difference (V′) is given by
V′=QC=1×10−9×2×10−3200×10−4×8.85×10−12=11.3 V
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