Question

an
electron stops under the action of an uniform electric field when
it has ran a distance of 90cm. If the velocity with which it enters
the field is v=2x10^6 i m/s, not considering the effect of
gravitational force, the magnitude and direction of the electric
field is:

a) 12.65 N/C i

b) 5.65 N/C -i

c) 1.60 N/C -i

d) 6.28 N/C i

Answer #1

The corret answer is 1) 12.65 N/c i

Work done = change in kinetic energy of the electron

So here, work done by the electic field = 0- 1/2 mv²

Where 0 is the final kinetic energy

m is the mass of electron = 9.1*10^{-31}Kg

V is the velocity of electron = 2*10^{6} m/s

So work done = 1/2 *
9.1*10^{-31}*(2*10^{6})²

W = 1.82*10^{-18} J

Here work done = force * displacement

And force = Eq

Where E is the electic field and q is the charge of electron=
1.6*10^{-19}C

Displacement = 90 cm = 0.9 m

So subtituting the values,

1.82*10^{-18} = E * 1.6*10^{-19} * 0.9

E = 1.82*10^{-18}/(1.6*10^{-19}*0.9)

E = 12.64 N/C

OR approximately elecric field = 12.65 N/

Since the electron is negative charged and elecric field travel from positive to negative charges, the direction of elecric field is towards + x axis (i).

So elecric field = 12.65i

(Please upvote if helpful)

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