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A cavity of radius r is made insideasolid sphere. The volume charge density of the remaining sphere is rho. An electron (charge e, mass m) is released inside the cavity from point P as shown in figure. The centre of sphere and center of cavity are separated by a distance a. The time after which the electron again touches the sphere is |
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Answer» `SQRT((6sqrt(2)repsilon_0m)/(epa))` `A=eE//m` (in backward direction).  and `E=(pa)/(3epsilon_(0))` or `A=(epa)/(2epsilonm)` Distance traveled by electron where it HITS the wall of cavity is `S=2rcostheta=(r)/(sqrt(s)` Hence using `s=(1)/(2)At^(2)` `t=sqrt((6sqrt(2)repsilon_(0)m)/(epa))` |
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