Question

A uniform electric field of 2.0×105 N/C is oriented in the negative x direction. How much work must be done to move an electron 3.0 m in the positive x direction, if it begins and ends at rest?

Answer #1

We know that work-done on a charged particle is given by:

W = F.d = F*d*cos A

F = Force on electron = q*E

Since charged particle is electron, which have negative charge, so force on electron will be in opposite direction of electric field

E = -2.0*10^5 N/C (-ve sign because electric field is in -ve x-axis)

F = q*E = (-1.6*10^-19)*(-2.0*10^5) = 3.2*10^-14 N in +ve x-axis

d = displacement of electron = 3.0 m

A = Angle between force and displacement = 0 deg

So,

W = F*d*cos A

W = 3.2*10^-14*3.0*cos 0 deg

**W = 9.6*10^-14 J**

**Let me know if you've any query.**

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