Animated Solution for Physics - Electrostatics: Two capacitors of capacitances $C$ and $2C$ are charged to potential differences $V$ and $2V$ respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is

Let's embark on this classic electrostatics problem by first understanding what we have before any connections are made. We are given two separate capacitors. The first capacitor has a capacitance of $C$ and is charged to a potential difference of $V$. The second capacitor is beefier, with a capacitance of $2C$, and is charged to a higher potential difference of $2V$.

To understand what happens when they interact, we must first determine the amount of charge each capacitor holds. Using the fundamental relation $Q=CV$, we can find the initial charges:

For the first capacitor: $Q_1=C\times V=CV$

For the second capacitor: $Q_2=2C\times 2V=4CV$

Opposite Polarity Connection

Now comes the critical part of the problem. The capacitors are connected in parallel, but with a twist: the positive terminal of one is connected to the negative terminal of the other.

Imagine two water tanks where you connect the high-pressure pipe of one to the low-pressure pipe of the other. They will fight each other! Similarly, the charges on the connected plates will neutralize each other. The net charge available to be distributed across the new parallel combination is the difference between their initial charges, not the sum.

$Q_{\text{net}}=Q_2-Q_1$ $Q_{\text{net}}=4CV-CV=3CV$

When capacitors are connected in parallel, they act as a single larger capacitor. The equivalent capacitance is simply the sum of the individual capacitances, regardless of how their polarities were connected.

Now, this combined system holds our net charge of $3CV$. Because they are in parallel, they must share a common potential difference. We can find this common potential by dividing the net charge by the equivalent capacitance:

$V_{\text{common}}=\frac{Q_{\text{net}}}{C_{\text{eq}}}$ $V_{\text{common}}=\frac{3CV}{3C}=V$

Finally, the question asks for the final energy of this configuration. The energy stored in any capacitor system is given by $U=\frac{1}{2}{C}_{\text{eq}}{V}_{\text{common}}^2$.

Substituting the values we just found:

$U_f=\frac{1}{2}(3C)(V{)}^2$ $U_f=\frac{3}{2}C{V}^2$

This elegant result shows how the energy redistributes. As a fun exercise, try calculating the initial total energy (which is $\frac{9}{2}C{V}^2$) and notice that $3C{V}^2$ of energy was lost as heat during the charge redistribution process!