An uniform electric field of strength `E` exists in a region. An electron of mass `m` enters a point A perpendicular to x - axis with velocity `V`. It moves through the electric field and exists at point `B`. The components of velocity at `B` are shown in (Fig. 3.156). At `B` the y - component of velocity remain uncharged.

Find the rate of work done by the field at `B`.

Options

A. `(2 m a^2 V^3)/(d^3)`
B. `( m a^2 V^3)/(2 d^3)`
C. `(4 m a^2 V^3)/(d^3)`
D. `(m a^2 V^3)/(d^3)`

Answer

( c)

Explanation

Electric field should be an negative x - direction because velocity along y - direction remains constant.
Accleration of electron along x - axis is `a_x = eE//m`.
Time taken to go from A to B is `t = d//V`.
Along x - axis
`a = (1)/(2) (a_x) t^2 = (1)/(2) (e E)/(m) (d/V)^(2)`
or `E = (2 m a V^2)/(e d^2)`
`v_1 = a_x t = (e E)/(m) d/V = (e)/(m) d/V (2 m a V^2)/(e d^2) = (2 a)/d V`
Velocity at `B is sqrt(V^2 + v_1^2) = Vsqrt (1 + ((2 a)/d)^2)`
Rate of work done by field at `B` is
`F v_1 = e E (2 a)/d V = (2 a e V)/d (2 m a V^2)/(e d^2) = (4 m a^2 V^3)/(d^3)`.
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