Question

# A parallel-plate capacitor is charged by being connected to a battery and is kept connected to the battery. The separation between the plates is then doubled. How does the electric field change? The charge on the plates? The total energy? Explain.

Solution

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#### Given

The parallel plates are kept connected to the battery. This means the applied voltage is fixed. Also, the distance $d$ between the two plates is doubled.

#### Required

The change in the electric field $E$, the charge $Q$ and the total energy $U$

#### Explanation

The electric field depends on the separated distance between the two plates and it is given by

$\begin{equation} E = \dfrac{V}{d} \end{equation}$

As shown by equation (1), the electric field is inversely proportional to the separated distance $d$ and the potential difference is constant. Hence, the electric field is $\textbf{halved}$ when the distance $d$ is doubled.

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