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Question

A capacitor is charged with a battery and then removed from the battery.
In this specially designed capacitor, we are able to make the plate size (area) larger without changing anything else. If the plate area is made larger after the capacitor has been disconnected, what will happen to the charge on the plates, the voltage across the plates, and the value of capacitance for this capacitor?

A
Capacitor Property Value after Plate Area is increased
Charge on Platessmaller
Voltage Across Platessmaller
Capacitance smaller
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B
Capacitor Property Value after Plate Area is increased
Charge on Plateslarger
Voltage Across Platessmaller
Capacitance large
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C
Capacitor Property Value after Plates Area is increased
Change on Platessame
Voltage Across Plateslarger
Capacitancesmaller
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D
Capacitor Property Value after Plates Area is increased
Change on Plates Same
Voltage Across Plates Same
Capacitance larger
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E
Capacitor Property Value after Plates Area is increased
Change on Plates same
Voltage Across Plates smaller
Capacitance larger
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Solution

The correct option is E
Capacitor Property Value after Plates Area is increased
Change on Plates same
Voltage Across Plates smaller
Capacitance larger

Since the capacitor is charged first and then removed from the battery and then made changes to, the charge Q on the capacitor remains the same.
The capacitance of the capacitor=C=Aϵ0d
Hence as the area of capacitor is increased, the capacitance of capacitor increases.
From the equation Q=CV
V1C
Hence the voltage across the plates decreases.
Correct answer is option E.

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