Answer

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**Hint:**The energy in a capacitor is given by the square of the charge divided by twice the capacitance. So, since the capacitance is constant, the energy will be proportional to square of the charge. So when the charge will be doubled, the energy becomes four times.

**Formula Used:**In this solution we will be using the following formula,

$E = \dfrac{{{Q^2}}}{{2C}}$

where $E$ is the energy of the condenser,

$Q$ is the charge on the condenser and

$C$ is the capacitance.

**Complete Step by Step Solution:**

The energy stored in a capacitor is the potential energy that is related to the voltage and charge on the capacitor. If the capacitance of a capacitor is given $C$ and the charge in it is given $Q$, where $V$ is the voltage across the plates, then the energy of the capacitor is given by the formula,

$E = \dfrac{1}{2}C{V^2}$

Now, for a capacitor, $Q = CV$

So we can replace the $V$ from the energy term by,

$V = \dfrac{Q}{C}$

So we get energy as,

$\Rightarrow E = \dfrac{1}{2}C\dfrac{{{Q^2}}}{{{C^2}}}$

So cancelling the $C$ from numerator and denominator, we get

$\Rightarrow E = \dfrac{{{Q^2}}}{{2C}}$

So from this formula we can see that the energy is proportional to the square of the charge.

Now in the question we are given that the charge in the capacitor is doubled. So the value of the capacitance is the same in both cases.

$\Rightarrow E \propto {Q^2}$

Therefore when the charge gets doubled,

$\Rightarrow E \propto {\left( {2Q} \right)^2}$

This gives us,

$\Rightarrow E \propto 4{Q^2}$

Therefore we can see that the energy becomes four times when the charge gets doubled.

**Thus, the correct answer will be option D.**

**Note:**A charged capacitor stores energy in the electric field between the plates. The electric field is built up when the capacitor is charged and when it gets disconnected, the energy remains in the field in between the plates.

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