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An electron moving with speed of \(5.0\times 10^8\)cm/sec is shot parallel to an electric field strength of \(1.0\times 10^3\)nt/coul arranged so as to retard its motion.
- How far will the electron travel in the field before coming (momentarily) to rest ?
- how much time will elapse?
- If the electric field ends abruptly after \(0.8\) cm, what fraction of its initial energy will the electron loose in traversing the field?
Solution
- The retardation of the electron due to electric field \(a=eE/m\).
- Initial velocity is given to be \(u=.0\times10^8\)cm/sec.
- The distance traveled \(s = \frac{u^2}{2a}\).
- Time taken, before it stops, is given by \(0=u-at\). Hence \(t=u/a\).
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