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(A) The electric field $E$ between two parallel plates is given by $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the distance bet

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$10^8\,m/s^2$

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(A) The electric field $E$ between two parallel plates is given by $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the distance between the plates.Given: $V = 50\,V$,$d = 5\,mm = 5 	imes 10^{-3}\,m$.$E = \frac{50}{5 	imes 10^{-3}} = 10^4\,V/m$.The force $F$ on the particle is $F = qE$.The acceleration $a$ is given by $a = \frac{F}{m} = \frac{qE}{m}$.Given: $q = 10^{-11}\,C$,$m = 10^{-15}\,kg$.$a = \frac{10^{-11} 	imes 10^4}{10^{-15}} = \frac{10^{-7}}{10^{-15}} = 10^8\,m/s^2$.

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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