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

 

Answers (1)

As we learnt in 

Parallel Plate Capacitor -

C=\frac{\epsilon _{0}A}{d}

- wherein

Area - A seperation between two plates.

 

 and

Energy Stored -

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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