Two parallel plates separated by a distance of $5\,mm$ are kept at a potential difference of $50\,V.$ A particle of mass ${10^{ - 15}}\,kg$ and charge ${10^{ - 11}}\,C$ enters in it with a velocity ${10^7}\,m/s.$ The acceleration of the particle will be
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(a) $a = \frac{{qE}}{m} = \frac{q}{m}\left( {\frac{V}{d}} \right)$$ = \frac{{{{10}^{ - 11}}}}{{{{10}^{ - 15}}}} \times \frac{{50}}{{5 \times {{10}^{ - 3}}}} = {10^8}\,m/se{c^2}$
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