Solution:

To calculate the work required to separate the plates of the capacitor, let’s analyze the situation step by step:

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1. Initial Setup

• The capacitor has:
  • Plate area =
    $A$ 
  • Initial separation =
    $d$ 
  • Initial potential difference =
    $V$ 
• After charging, the battery is disconnected, so the charge
  $Q$ 

  on the plates remains constant.

The charge stored in the capacitor is given by:

 

$$Q = C V$$

 

where

$C$

is the capacitance:

 

$$C = \frac{\varepsilon_0 A}{d}$$

 

Thus, the charge is:

 

$$Q = \frac{\varepsilon_0 A}{d} V$$

 

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2. Energy Stored in the Capacitor

The energy stored in the capacitor is:

 

$$U = \frac{1}{2} C V^2$$

 

Substituting

$C = \frac{\varepsilon_0 A}{d}$

:

 

$$U = \frac{1}{2} \cdot \frac{\varepsilon_0 A}{d} \cdot V^2$$

 

Initially, the energy is:

 

$$U_{\text{initial}} = \frac{\varepsilon_0 A V^2}{2d}$$

 

When the separation is increased to

$2d$

, the capacitance decreases to:

 

$$C_{\text{new}} = \frac{\varepsilon_0 A}{2d}$$

 

The energy becomes:

 

$$U_{\text{final}} = \frac{1}{2} C_{\text{new}} V_{\text{new}}^2$$

 

Since the charge

$Q$

is constant and

$Q = C_{\text{new}} V_{\text{new}}$

:

 

$$V_{\text{new}} = \frac{Q}{C_{\text{new}}} = \frac{\frac{\varepsilon_0 A}{d} V}{\frac{\varepsilon_0 A}{2d}} = 2V$$

 

Substitute

$C_{\text{new}}$

and

$V_{\text{new}}$

:

 

$$U_{\text{final}} = \frac{1}{2} \cdot \frac{\varepsilon_0 A}{2d} \cdot (2V)^2$$

 

$$U_{\text{final}} = \frac{\varepsilon_0 A V^2}{2d}$$

 

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3. Work Done to Separate the Plates

The work done to separate the plates is equal to the increase in energy:

 

$$W = U_{\text{final}} - U_{\text{initial}}$$

 

Substitute the values:

 

$$W = \frac{\varepsilon_0 A V^2}{2d} - \frac{\varepsilon_0 A V^2}{2d}$$

 

$$W = \frac{\varepsilon_0 A V^2}{2d}$$

 

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Final Answer:

The work required to separate the plates is:

 

$$W = \frac{\varepsilon_0 A V^2}{2d}$$