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(C) The electron is projected with kinetic energy $K = \frac{1}{2}mu^2$ at an angle $	heta = 45^\circ$ in a uniform electric field $E$ directed upwar

Vedclass · Accepted Answer

$\frac{K}{2qd}$

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(C) The electron is projected with kinetic energy $K = \frac{1}{2}mu^2$ at an angle $	heta = 45^\circ$ in a uniform electric field $E$ directed upwards. The force on the electron is $F = qE$ downwards, so the acceleration is $a = \frac{qE}{m}$ downwards.The maximum vertical height $H$ reached by the electron is given by the formula for projectile motion: $H = \frac{u^2 \sin^2 	heta}{2a}$.Substituting the values $H = d$, $	heta = 45^\circ$, and $a = \frac{qE}{m}$:$d = \frac{u^2 \sin^2 45^\circ}{2(qE/m)}$Since $\sin 45^\circ = \frac{1}{\sqrt{2}}$, we have $\sin^2 45^\circ = \frac{1}{2}$.$d = \frac{u^2 (1/2)}{2qE/m} = \frac{mu^2}{4qE}$Given $K = \frac{1}{2}mu^2$, we can write $mu^2 = 2K$.Substituting this into the equation:$d = \frac{2K}{4qE} = \frac{K}{2qE}$Therefore, $E = \frac{K}{2qd}$.For the electron to strike the upper plate, the electric field must be such that the maximum height reached is at least $d$. Thus, $E$ must be greater than $\frac{K}{2qd}$.

As shown in the figure, an electron is projected with kinetic energy $K$ at an angle of $\theta = 45^\circ$ between two charged plates. The electron strikes the upper plate. The value of the electric field $E$ must be greater than:

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