Work done - moving a charge in an electric field

[cutesteph]

work done -- moving a charge in an electric field

Homework Statement

Q=10uC from origin to (3m,pi/4,pi/2) E=10r unit r + 5/(rsin(theta) unit phi V/m

answer is -475uJ

I found this problem in a reference book that gave answers, this problem is similar to a homework problem.

Homework Equations

w = -q ∫ E*dl or w = 1/2 integral (εE^2) dv

The Attempt at a Solution

w= -10x10-6 ∫ 10rdr + 5d(phi)

 

[berkeman]

Homework Statement

Q=10uC from origin to (3m,pi/4,pi/2) E=10r unit r + 5/(rsin(theta) unit phi V/m

answer is -475uJ

I found this problem in a reference book that gave answers, this problem is similar to a homework problem.

Homework Equations

w = -q ∫ E*dl or w = 1/2 integral (εE^2) dv

The Attempt at a Solution

w= -10x10-6 ∫ 10rdr + 5d(phi)

Could you please clarify the question a bit? When you give the electric field, is that in spherical or rectangular coordinates? Would it be possible to scan the problem and upload it into this thread?

 

[cutesteph]

The problem is in spherical coordinates.

My work is w = -10x10^-6 ( 20( sqrt(exp(1) -1) +10 pi/2 = 5.736 uJ , which is incorrect from the answer in the book.

 

[cutesteph]

I am not sure what I am doing wrong, dL = dr (unit r) + r dtheta (unit theta) + r sin theta d phi (unit phi) .
E=10r (unit r) + 5/[(rsin(theta)] (unit phi) V/m

so integral of work = q * (10r dr + r phi dphi )

from origin 0,0,0 to 3, pi/4, pi /2 so r goes 0 to 2 theta goes 0 to pi/4 and phi goes 0 to pi/2

 

[rude man]

Are you sure about the second term in the E field expression? Because that implies an infinite E field at the origin.

Anyway, two hints:

1. the E field is conservative so what does that tell you about the path of integration?

2. You can split the E field into two fields, corresponding to the two terms, and use superposition. Then go back to hint #1.

I assume theta is the angle with the z axis and phi is the azimuth angle, and that you're giving the final position as (r, theta, phi).

(My answer is close to -475 uJ but not quite.)

 

[cutesteph]

Since E is conservative, the path of integration does not matter.

 

[rude man]

Since E is conservative, the path of integration does not matter.

Right. So what path would you take for E dot dl?