Question
Through what potential difference must an electron be accelerated, starting from rest, to acquire a speed of $0.99 c$ ?
Step 1
99c$ (final velocity) $u = 0$ (initial velocity) $c = 3.00 \times 10^8$ m/s (speed of light) Plugging in the values: $0.99c = \frac{0 + v'}{1 + \frac{0 \times v'}{(3.00 \times 10^8)^2}}$ Solving for $v'$: $0.99 = \frac{v'}{1 + 0}$ $v' = 0.99c$ Show more…
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