Question

a parallel plate capacitor has area of each plate a the separation between the plates is d. it is charged to a potential v and then disconnected from the battery. and plates are pulled to 3times the initial separation then work done to separate the plate is​

Answer 1

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Answer 2

Answer:

$\sf W=\dfrac{A\: \epsilon_0}{d}\times V^2$

Explanation:

Given:

• A parallel plate capacitor of area A and d as the distance of separation between the plates
• The plate is charged to a potential V and is pulled apart to 3 times the initial separation.

To Find:

• The work done to separate the plates

Solution:

The initial charge on the capacitor is given by,

Q = C V

where C is the capacitance and V is the voltage

$\sf C = \dfrac{A\: \epsilon_0}{d}$

where A is the area and d is the distance of the plates.

By given the new distance is 3 times the initial one,

Hence new capacitance (C') is given by,

$\sf C' = \dfrac{A\: \epsilon_0}{3d}=\dfrac{1}{3} \times C$

We know that work done on a parallel plate capacitor is given by,

Work done = Final Energy - Initial energy

Energy stored in a capacitor (U) is given by,

$\sf U=\dfrac{1}{2} \: \dfrac{Q^2}{C}$

Hence,

$\sf W=\bigg(\dfrac{Q^2}{2C'} -\dfrac{Q^2}{2C} \bigg)$

Substitute the value of C',

$\sf W=\bigg(\dfrac{Q^2}{2\times C/3} -\dfrac{Q^2}{2C} \bigg)$

$\sf W=\bigg(\dfrac{3Q^2}{2 C} -\dfrac{Q^2}{2C} \bigg)$

$\sf W=\dfrac{2Q^2}{2C} =\dfrac{Q^2}{C}$

We know,

Q = CV

Hence,

$\sf W =\dfrac{C^2\: V^2}{C}$

$\sf W=CV^2$

Substitute value of C,

$\sf W=\dfrac{A\: \epsilon_0}{d}\times V^2$

This is the work done to separate the plates.