An electron of charge e in rest in an electric field between two plate

1.	An electron of charge e in rest in an electric field between two plates separated by distance d and with potential difference V, the force experienced by the charge is given by(a)eVd (b)edV (c)dqV (d)Vd
Answer» Given: An ELECTRON of charge e in rest in an ELECTRIC field between two PLATES separated by distance d and with potential difference V. To find: Force experienced by the charge ? Calculation: The general expression for force experienced by a STATIC charge q is : $\therefore \: F = q \times E$ Now, Electrostatic field intensity can be expressed as : $\implies \: F = q \times ( - \dfrac{ \partial V}{ \partial \: r} )$ $\implies \: \| F \| = q \times \dfrac{ \partial V}{ \partial \: r}$ $\implies \: \| F \| = q \times \dfrac{ V}{ d }$ $\implies \: \| F \| = e\times \dfrac{ V}{ d }$ $\implies \: \| F \| = \dfrac{eV}{ d }$ So, final answer is : $\boxed{\: \bf \| F \| = \dfrac{eV}{ d } }$