Special theory of relativity > The transformation of electric and magnetic fields > Problem 14.3.10

Statement

$14.3.10.$ How many times will the potential difference and capacitance of a long cylindrical capacitor change when it moves at a speed $\beta c$ along the axis?

Solution

There are two ways to solve this problem; we will see both.

Because charge is a conserved quantity, we can say that $Q = C V \rightarrow C V = \text{constant}$ for all frames.
Suppose that in the frame of the capacitor, the capacitor has a height $h$ and a surface area $S$; then the capacitance in the moving frame is:

\begin{equation}
C = \frac{\varepsilon_0 S}{h}
\end{equation}

In the Earth frame, the capacitor is moving with velocity $\beta c$ along its axis; then in the Earth frame the surface area of the capacitor will be the same,
but the height will be:

\begin{equation}
h' = h \sqrt{1-\beta^2}
\end{equation}

Thus, the capacitance will be:

\begin{equation}
C' = \frac{\varepsilon_0 S}{h'} = \frac{\varepsilon_0 S}{h \sqrt{1-\beta^2}} = \frac{C}{\sqrt{1-\beta^2}}
\end{equation}

Using the conserved quantity $Q$, we can obtain the voltage $V$:

\begin{equation}
V'C' = V C \rightarrow V' = V \sqrt{1-\beta^2}
\end{equation}

On the other hand, we can use the equation for the voltage of the capacitor $V = E h$. Using it, we see that because the distance between plates in the Earth frame will be $\sqrt{1-\beta^2}$ times smaller, and the field does not change between frames,
the voltage in the Earth frame will be $\sqrt{1-\beta^2}$ times smaller.

Answer

\begin{equation}
C' = \frac{C}{\sqrt{1-\beta^2}}
\end{equation}

\begin{equation}
V' = V \sqrt{1-\beta^2}
\end{equation}