Solution:

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

$\frac{1}{2}$

.