Question:

A capacitor is connected to a battery with a potential difference of 12 volts. The capacitor has a capacitance of 6 microfarads. The battery is then disconnected and the plates of the capacitor are moved farther apart until the separation between them is doubled. Calculate the new electric potential energy stored in the capacitor.

Answer:

To calculate the new electric potential energy stored in the capacitor, we need to use the formula:

where:

• $U$ is the electric potential energy stored in the capacitor,
• $C$ is the capacitance of the capacitor,
• $V$ is the voltage across the capacitor.

We are given that the original voltage across the capacitor ($V_{\text{original}}$) is 12 volts and the original capacitance ($C_{\text{original}}$) is 6 microfarads.

First, let's calculate the original electric potential energy stored in the capacitor ($U_{\text{original}}$) using the given information:

Substituting the values, we get:

Simplifying, we find:

Now, we are asked to find the new electric potential energy ($U_{\text{new}}$) after the plates of the capacitor are moved farther apart until the separation between them is doubled.

When the separation between the plates is doubled, the capacitance of the capacitor becomes half of its original value ($C_{\text{new}}$ = $\frac{1}{2} C_{\text{original}}$). We can use the formula for electric potential energy to calculate $U_{\text{new}}$ using $C_{\text{new}}$ and the original voltage ($V_{\text{original}}$):

Substituting the values, we have:

Simplifying further, we get:

Simplifying, we find:

Therefore, the new electric potential energy stored in the capacitor after doubling the separation between the plates is 0.162 joules.