Правка разделов «Statement», «Solution», «Answer»

Alexphysics правка от
правка #18986 предыдущая #18985 ← раньше
@@ -1,7 +1,11 @@
### Statement
−$14.3.22.$ [Insert the problem statement]
+$14.3.22.$
+Thickness of a fixed flat capacitor h, leakage current density j. Initial surface
+charge density σ. How will the electric field inside the capacitor change when
+it moves at a speed βc parallel to the plates?
+
### Solution
Rest frame S'
@@ -28,9 +32,7 @@Solution
For a boost in the x-direction, the fields transform according to:
$\begin{aligned}
−E_x &= E'_x, \quad & E_y &= \gamma(E'_y + v B'_z), \quad & E_z &= \gamma(E'_z - v B'_y),\\[4pt]
−B_x &= B'_x, \quad & B_y &= \gamma\!\left(B'_y - \frac{v}{c^2}E'_z\right), \quad & B_z &= \gamma\!\left(B'_z + \frac{v}{c^2}E'_y\right),
−\end{aligned}$
+E_x &= E'_x, \quad & E_y &= \gamma(E'_y + v B'_z), \quad & E_z &= \gamma(E'_z - v B'_y) & B_x = B'_x, \quad B_y = \gamma\left(B'_y - \frac{v}{c^2}E'_z\right), \quad B_z = \gamma\left(B'_z + \frac{v}{c^2}E'_y\right)\end{aligned}$
with
@@ -55,4 +57,4 @@Solution
#### Answer
−[Insert a concise answer or boxed result]
+$\boxed{E_z = \gamma E_0 = \frac{E_0}{\sqrt{1-\beta^2}}}$