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A parallel-plate capacitor is made of two conducting plates of area A separated by a distance d. The capacitor carries a charge Q and is initially connected to a battery that maintains a constant potential difference between the plates. The battery is then disconnected from the plates and the separation between the plates is doubled. The electric potential energy stored in the capacitor will:

Respuesta :

Answer:

Stored potential energy will increase.

Explanation:

Given that

Area = A

Capacitance =C

Charge =Q

We  know that stored potential energy given as

[tex]U=\dfrac{1}{2}CV^2[/tex]

[tex]U=\dfrac{1}{2C}Q^2[/tex]

[tex]C=\dfrac{\varepsilon _oA}{d}[/tex]

So when distance (d) between plates become double then C will become C/2.

So new value of capacitance C'=C/2

[tex]U'=\dfrac{1}{2C'}Q^2[/tex]          (Charge will remain same)

[tex]U'=\dfrac{1}{2\dfrac{C}{2}}Q^2[/tex]

[tex]U'=2\dfrac{1}{2{C}}Q^2[/tex]

U' =2 U

It means that stored potential energy will increase.

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