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
• B. Decrease
• C. Remain the same
• D. Become zero

Answer 1

Answer: (A)

Increase

Let, the capacitance of the air capacitor is $C_0$

When the capacitor is dipped into oil tank, the capacitance becomes $C^{\prime }=K{C}_0$ where dielectric constant of oil medium.

Now charge on capacitor is $q^{\prime }=C^{\prime }{V}^{\prime }$

When oil pump out , the charge is $q=C_{0}V$

As battery is not connected, so charge remains constant.

Thus, $q^{\prime }=q$ or $C^{\prime }{V}^{\prime }=C_{0}V$ or $K{C}_0{V}^{\prime }=C_{0}V$

or $V=K{V}^{\prime }$ or $V>V'

If d be the distance between capacitor plates, the electric field $V=Ed$

As $V\propto E$, the electric field will increase (as $V>V^{\prime }$) when oil is pumped out.