Question

A uniform electric field of magnitude 4.1 105 N/C points in the positive x-direction. Find the change in electric potential energy of a +5.1 µC charge as it moves fromthe origin to each of the points given below.
(a) (0, 5.7 m)
J

(b) (5.7 m, 0)
J

(c) (5.7 m, 5.7 m)
J

Answer 1

given,

electric field E=4.1*10^5 N/C is along x-direction

charge q=3.8uC

a)

(x, y)=(0,5.1m)

change in P.E(dU) is workdone=F.s

dU=(qE)s*cos(theta)

here, theta=90 degrees

dU=(qE)s*cos(90)

=0

b)

(x, y)=(5.1m, 0)

change in P.E(dU) is workdone=F.s

dU=(qE)s*cos(theta)

here, theta=0 degrees

dU=(qE)s*cos(0)

=(3.8*10^-6*4.1*10^5)*5.1*1

=7.946 J

c)

(x, y)=(5.1m, 5.1m)

change in P.E(dU) is workdone=F.s

dU=(qE)s*cos(theta) -----along x-axis

here, theta=45 degrees

dU=(qE)s*cos(45)

=(3.8*10^-6*4.1*10^5)*[sqrt(2)*5.1]*cos(45)

=7.946 J

and

dU=(qE)s*cos(theta) -----along y-axis

here, theta=90 degrees

dU=(qE)s*cos(90)

=0