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(B) The force $F$ on an electron in an electric field $E$ is given by $F = qE$.The electric field between two parallel plates is $E = \frac{V}{d}$,whe

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$8 	imes 10^{-14}\,N$

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(B) The force $F$ on an electron in an electric field $E$ is given by $F = qE$.The electric field between two parallel plates is $E = \frac{V}{d}$,where $V$ is the potential difference and $d$ is the separation distance.Given: $q = 1.6 	imes 10^{-19}\,C$,$V = 1000\,V$,and $d = 2\,mm = 2 	imes 10^{-3}\,m$.Substituting these values into the formula:$F = q \left( \frac{V}{d} ight) = (1.6 	imes 10^{-19}) \left( \frac{1000}{2 	imes 10^{-3}} ight)$$F = (1.6 	imes 10^{-19}) 	imes (0.5 	imes 10^{6})$$F = 0.8 	imes 10^{-13}\,N = 8 	imes 10^{-14}\,N$.

An electron enters between two horizontal plates separated by $2\,mm$ and having a potential difference of $1000\,V$. The force on the electron is:

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