The plates of a parallel plate capacitor are charged with a battery so that the plates of the capacitor have acquired the P.D. equal to e.m.f of the battery. The ratio of the work done by the battery and the energy stored in capacitor is

Options

A. `1:2`
B. `1:1`
C. `1:2`
D. `1:4`

Answer

To solve the problem, we need to find the ratio of the work done by the battery to the energy stored in the capacitor. Let's break it down step by step.

### Step 1: Understand the Work Done by the Battery
When a capacitor is charged by a battery, the work done by the battery (W) can be expressed as:
\[ W = Q \cdot V \]
where \(Q\) is the charge on the capacitor and \(V\) is the potential difference (P.D.) across the capacitor, which is equal to the e.m.f of the battery.

### Step 2: Relate Charge to Capacitance
The charge \(Q\) on the capacitor is related to its capacitance \(C\) and the voltage \(V\) across it:
\[ Q = C \cdot V \]

### Step 3: Substitute Charge into Work Done Equation
Substituting the expression for \(Q\) into the work done equation gives:
\[ W = (C \cdot V) \cdot V = C \cdot V^2 \]

### Step 4: Calculate the Energy Stored in the Capacitor
The energy \(U\) stored in a capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \]

### Step 5: Find the Ratio of Work Done to Energy Stored
Now, we can find the ratio of the work done by the battery to the energy stored in the capacitor:
\[ \text{Ratio} = \frac{W}{U} = \frac{C V^2}{\frac{1}{2} C V^2} \]

### Step 6: Simplify the Ratio
This simplifies to:
\[ \text{Ratio} = \frac{C V^2}{\frac{1}{2} C V^2} = \frac{1}{\frac{1}{2}} = 2 \]

### Conclusion
Thus, the ratio of the work done by the battery to the energy stored in the capacitor is:
\[ \text{Ratio} = 2:1 \]