A capacitor is charged until its stored energy is 4.00 J. A second capacitor is then connected to it in parallel. (a) If the charge distributes equally, what is the total energy stored in the electric fields? (b) Where did the missing energy go?

• The energy stored in the capacitor$U_0=4.0\mathrm{J}$
• The charges $q_1$and $q_2$are equally distributed when both the capacitors are connected. That is,$q_1=q_2=\frac{Q_0}{2}$ .

The charges in the capacitor are conserved, we cannot create new charges or destroy the existing charges. The charges can move from one plate to another. The capacitors with the charges have the energy stored in them. The energy can be written in terms of charge and capacitance of the capacitors.

Here, we use the charge conservation in the capacitors and the energy stored in the capacitor. From this, we can find out the energy stored in the given capacitor arrangement.

Formulae:

According to charge conservation law, $Q=q_1+q_2$ …(i)

The energy stored in the electric field, $U=\frac{Q^2}{2C}$ …(ii)

The energy stored within the capacitors can be given using equation (ii) as follows:

$$\begin{gathered} U_0=\frac{Q_0^2}{2C} \\ =4.0\mathrm{J} \end{gathered}$$

Suppose$U_1\mathrm{and}{U}_2$ are the energies stored in first and second capacitor, when they are connected, and the charges are distributed equally. So, we can write their individual stored energy using equation (i) as follows:

$$\begin{gathered} U_1=\frac{{\left({\displaystyle \frac{Q_0}{2}}\right)}^2}{2C} \\ =\frac{Q_0^2}{8C} \\ =\frac{U_0}{4} \\ =\frac{4\mathrm{J}}{4} \\ =1.00\mathrm{J} \end{gathered}$$

Similarly, for$U_2$we get the stored energy value as due to having similar charges with capacitor 1.

Thus, the total energy stored in the electric field is given as:

Hence, the value of the energy stored is $2.0\mathrm{J}$.

Now, the energy loss in this arrangement of the capacitorsis given by:

$$\begin{gathered} U_0-U=\left(4-2\right)\mathrm{J} \\ =2.0\mathrm{J} \end{gathered}$$.

Hence, the value of the energy is lost in the overcoming resistance from the wires is $2.0\mathrm{J}$.