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+### Statement
+
+$14.3.9.$ [Insert the problem statement]
+
+### 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 \( \frac{1}{\sqrt{1 - \beta^2}} \).
+
+### Answer
+
+- a) The potential difference increases by a factor of \( \frac{1}{\sqrt{1 - \beta^2}} \)
+- b) The potential difference decreases by a factor of \( \frac{1}{\sqrt{1 - \beta^2}} \)
+
+
+#### Answer
+
+[Insert a concise answer or boxed result]