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valkas [14]
3 years ago
12

There's an electric field in some region of space that doesn't change with position. An electron starts moving with a speed of 2

.0 × 107 m/s in a direction opposite to the field. Its speed increases to 4.0 × 107 m/s over a distance of 1.2 cm. What is the magnitude of the electric field?
Physics
1 answer:
tangare [24]3 years ago
5 0

Answer:

Explanation:

Given

speed of Electron u=2\times 10^7\ m/s

final speed of Electron v=4\times 10^7\ m/s

distance traveled d=1.2\ cm

using equation of motion

v^2-u^2=2as

where v=Final velocity

u=initial velocity

a=acceleration

s=displacement

(4\times 10^7)^2-(2\times 10^7)^2=2\times a\times 1.2\times 10^{-2}

a=5\times 10^{16}\ m/s^2

acceleration is given by a=\frac{qE}{m}

where q=charge of electron

m=mass of electron

E=electric Field strength

5\times 10^{16}=\frac{1.6\times 10^{-19}\cdot E}{9.1\times 10^{-31}}

E=248.3\ kN/C                

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The value is  C = 30729\  c

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From the question we are told that

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