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An electric field of 1×106N/C acts on an electron, resulting in an acceleration of 1.8×1017m/s2 for a short time. What is the magnitude of the radiative electric field observed at a location a distance of 2cm away along a line perpendicular to the direction of the acceleration?

Short Answer

Expert verified

The radiative electric field is 1.44×10-7N/C.

Step by step solution

01

Given data

Applied electric field:

E=1×106N/C

Acceleration of the electron:

a=1.8×1017m/s2

Distance of measurement of radiative electric field

r=2cm=2·1cm×1m100cm=0.02m

02

Determine the concept of electric field

The radiative electric field of a particle of charge and acceleration is:

E=14πε0qac2r ……. (i)

Here, ε0is the permittivity of free space with value 14πε0=9×109N.m2/C2

Here, Eis the speed of light in vacuum with charge c=3×108m/s.

03

Determine the radiative electric field

From equation (i), the radiative electric field for the electron with charge 1.6×10-19Cis calculated as:

E=9×109N.m2/C2×1.6×10-19C×1.8×1017m/s23×108m/s2×0.02m=1.44×10-7N/C

Thus, the field is 1.44×10-7N/C.

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