An electron (charge - 1.6 ´ 10 - 19 C) moves on a path perpendicular to the direction of a uniform electric field of strength 3.0 N/C. How much work is done on the electron as it moves 15 cm?

Explanation:

The charge on the electron is

q=1.6×10^-19C

It moves in a path perpendicular to the electric field

Given that the uniform electric field is

E=3N/C.

We need to calculate Work

Distance moved is

d=15cm=0.15m

W=q ∫E.ds

Thus, the angle θ between the path and electric field is 90°, and the dot product E.ds is zero

E.ds= Eds Cos θ= Edscos90

Cos 90=0

Then E.ds=0

Now,

W=q ∫E.ds

W=q × 0

W=0J

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explanation: