http://www.webassign.net/plemalg/images/1-pre-04.gif (a) Let q= +5 µC. How much

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http://www.webassign.net/plemalg/images/1-pre-04.gif

(a) Let q= +5 µC. How much work is done by you in moving this charge 4.2 m in the positive x-direction?
Got this answer: -630 J

(b) What is the potential difference?


(c) Let q= +5 µC. How much work is done by you in moving this charge 4.2 m in the negative x-direction?
God this answer: 630 J

(d) What is the potential difference?


(e) Let q= -5 µC. How much work is done by you in moving this charge 4.2 m in the positive y-direction?


(f) What is the potential difference?


(g) Let q= -5 µC. How much work is done by you in moving this charge 4.2 m at an angle of 30° to the electric field?


(h) What is the potential difference?



Please explain how you get the answers. Will rate the quickest and most accurate response.

I know F= q*E and work=F*d, I'm just not getting correct answers.

Explanation / Answer

b) potential can be found two ways: either v=Er or u = qv. They will both give you the same answer.

Lets use u = qv. You already have the energy, u, which is 630 J. You have q, so solving for v we get:

-630 J = (5e-6 C)v
v = -1.26e8 V.

d) same answer, 1.28e8 V but positive

e) Zero. you are moving perpendicular tot he electric field line, which means you are moving ALONG equipotential lines, so there is no change in potential, therefore no work. Also,m there is no force that will act against your motion, so no work either.

f) ZERO

g) W=Fd but with a dot product, so really Fdcos() or qEdcos()

We already have qEd, which is 630 J, so 630cos(30) = 545.6 J. The answer should be positive as well.

h)

u = qv

545.6 J = ( -5e-6 )v

v = -1.091e8 V

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