In a parallel plate capacitor made by stacking

$n$

equally spaced plates connected alternatively:

Key Points:

1. The plates connected alternatively form a series of capacitors.
2. If there are
   $n$ 

   plates, the number of gaps (capacitors in series) between them is $n - 1$ 

   .

For capacitors in series:

The total capacitance

$C_{\text{total}}$

is given by:

$$\frac{1}{C_{\text{total}}} = \frac{1}{C} + \frac{1}{C} + \dots + \frac{1}{C} \, \, (\text{n - 1 times})$$

$$\frac{1}{C_{\text{total}}} = \frac{n - 1}{C}$$

$$C_{\text{total}} = \frac{C}{n - 1}$$

However, because the plates are connected alternatively, these effectively act as

$n - 1$

capacitors in parallel.

For capacitors in parallel:

The equivalent capacitance is:

$$C_{\text{eq}} = (n - 1)C$$

Thus, the resultant capacitance is

$(n - 1)C$

.

To find the ratio of the energy stored in the capacitor to the work done by the battery, let's break it down step by step:

1. Energy Stored in the Capacitor

The energy

$U$

stored in a capacitor is given by:

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

where

$C$

is the capacitance, and

$V$

is the voltage across the capacitor (equal to the EMF of the battery once fully charged).

2. Work Done by the Battery

The work done by the battery

$W$

is equal to the total charge delivered multiplied by the voltage:

$$W = Q \cdot V$$

The charge

$Q$

stored in the capacitor is:

$$Q = C V$$

So, the work done becomes:

$$W = C V \cdot V = C V^2$$

3. Ratio of Energy Stored to Work Done

Now, the ratio of the energy stored in the capacitor to the work done by the battery is:

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

Final Answer:

The ratio is

$\frac{1}{2}$

.

Let’s solve step by step why the energy stored becomes

$2U$

:

1. Initial Setup

• Capacitance of the capacitor: 

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

  • 
    $A$ 

    = area of the plates,

  • 
    $d$ 

    = distance between the plates,

  • 
    $\varepsilon_0$ 

    = permittivity of free space.

• When the capacitor is charged by a battery to a voltage
  $V$ 

  , the charge stored is: 

  $$Q = CV$$The energy stored in the capacitor is:

   

  $$U = \frac{1}{2} C V^2$$ 

2. When the Distance is Doubled

After the battery is removed, the charge

$Q$

on the capacitor remains constant because there’s no external connection. However, the capacitance changes due to the increased plate separation.

• New capacitance: 

  $$C' = \frac{\varepsilon_0 A}{2d}$$ 

• Energy stored in a capacitor is given by: 

  $$U' = \frac{Q^2}{2C'}$$Substitute

  $C' = \frac{\varepsilon_0 A}{2d}$:

   

  $$U' = \frac{Q^2}{2 \cdot \frac{\varepsilon_0 A}{2d}} = \frac{Q^2 \cdot 2d}{2 \varepsilon_0 A}$$Simplify:

   

  $$U' = \frac{Q^2 d}{\varepsilon_0 A}$$ 

3. Relating $U'$

and $U$

From the initial setup:

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

Substitute

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

and

$Q = CV$

:

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

From the doubling of plate separation:

$$U' = 2 \cdot \frac{Q^2}{2C} = 2U$$

Final Answer:

The energy stored in the capacitor after doubling the separation is:

$$\boxed{2U}$$

On inserting dielectric Capacitance of the system increases so more charge is given by the battery.