Solution

Problem

$14.3.9.$ By what factor do the potential difference and capacitance of a parallel-plate capacitor change when it moves with velocity $\beta c$:
a) along the direction of the plates?
b) perpendicular to the plates?

Solution

The charge on the plates, $Q$, remains constant in all reference frames because charge is an invariant quantity in the theory of relativity. That is,

$$Q = CU \quad \Longleftrightarrow \quad \text{Invariant}$$

Let the Lorentz factor be denoted as:

$$\gamma = \frac{1}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} = \frac{1}{\sqrt{1 - \beta^2}}$$

a) Motion Along the Plates

In this case, relativistic effects manifest as follows:

• The distance between the plates $d$ remains unchanged, since length contraction only affects dimensions along the direction of motion.

• The area of the plates decreases due to contraction in one dimension:

  $$S' = \frac{S}{\gamma}$$

• The capacitance becomes:

  $$C' = \frac{\varepsilon_0 S'}{d} = \frac{\varepsilon_0 \frac{S}{\gamma}}{d} = \frac{C}{\gamma}$$

Using charge invariance $Q = C U = C' U'$, we find that the potential difference increases by a factor of $\gamma$:

$$U' = \gamma U = \frac{1}{\sqrt{1 - \beta^2}} U$$

Thus, the potential difference increases by a factor of $\frac{1}{\sqrt{1 - \beta^2}}$.

b) Motion Perpendicular to the Plates

Here, the relativistic effects are:

• The distance between the plates contracts:

  $$d' = \frac{d}{\gamma}$$

• The area of the plates remains the same:

  $$S' = S$$

• The capacitance becomes:

  $$C' = \frac{\varepsilon_0 S}{d'} = \frac{\varepsilon_0 S}{\frac{d}{\gamma}} = \gamma C$$

Again using charge invariance $Q = C U = C' U'$, the potential difference decreases by a factor of $\gamma$:

$$U' = \frac{U}{\gamma}$$

So, the potential difference decreases by a factor of $\frac{1}{\sqrt{1 - \beta^2}}$.