# A parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles, then Option 1) the charge on the capacitor increases Option 2) the voltage across the plates decreases Option 3) the capacitance increases Option 4) the electrostatic energy stored in the capacitor increases

As we learnt in

Parallel Plate Capacitor -

$\dpi{100} C=\frac{\epsilon _{0}A}{d}$

- wherein

Area - A seperation between two plates.

and

Energy Stored -

$\dpi{100} U=\frac{1}{2}CV^{2}=\frac{1}{2}QV=\frac{Q^{2}}{2C}$

-

If we move the plates apart, its capacitance, $C = \frac{\epsilon _{0}A}{d}$   will decrease.

It will have impact on its potential which will decrease but charge will remain same.

$Energy\ = \frac{1}{2}CV^{2} = \frac{Q^{2}}{2C}$    will increase as Q is constant and C decreases.

Option 1)

the charge on the capacitor increases

This option is incorrect

Option 2)

the voltage across the plates decreases

This option is incorrect

Option 3)

the capacitance increases

This option is incorrect

Option 4)

the electrostatic energy stored in the capacitor increases

This option is correct

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