1.

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 electric field.

Answer»

`(2 m aV^2)/(e d^2)`
`(m a V^2)/(e d^2)`
`(m aV^2)/(2 e d^2)`
`(2 m aV^3)/(e d^3)`

Solution : 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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