Question

an electron stops under the action of an uniform electric field when it has ran a...

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

Homework Answers

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*10-31Kg

V is the velocity of electron = 2*106 m/s

So work done = 1/2 * 9.1*10-31*(2*106

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*10-19C

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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