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Question

An air capacitor is charged with an amount of charge q and dipped into an oil tank. From the oil tank If the oil is pumped out, the electric field between the plates of capacitor will be:

A
Increase
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B
Decrease
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C
Remain the same
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D
Become zero
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Solution

The correct option is C Increase
Let, the capacitance of the air capacitor is C0
When the capacitor is dipped into oil tank, the capacitance becomes C=KC0 where dielectric constant of oil medium.
Now charge on capacitor is q=CV
When oil pump out , the charge is q=C0V
As battery is not connected, so charge remains constant.
Thus, q=q or CV=C0V or KC0V=C0V
or V=KV or $V>V'
If d be the distance between capacitor plates, the electric field V=Ed
As VE, the electric field will increase (as V>V) when oil is pumped out.

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