Work on charge from outside to inside of sphere

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  • #1
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Homework Statement


A solid non-conducting sphere of radius R = 1.12m.
The sphere posses a total charge Qtot spread uniformally throughout its volume.

a) derive equations for electric field for
1) 0<r<R
2) r>R
result in terms of r R and Q

b) Derive an equation that gives the work needed for you to bring a point charge "q" from infinitely far away to inside the sphere at r = R/2

Answer is W = 11/8 (kQ/R) q

Homework Equations



for 0<r<R
E1 = KQR/r^3

for r>R
E2 = KQ/R^2

The Attempt at a Solution



q∫KQR/r^3 + q∫KQ/R^2
 

Answers and Replies

  • #2
collinsmark
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Homework Statement


A solid non-conducting sphere of radius R = 1.12m.
The sphere posses a total charge Qtot spread uniformally throughout its volume.

a) derive equations for electric field for
1) 0<r<R
2) r>R
result in terms of r R and Q

b) Derive an equation that gives the work needed for you to bring a point charge "q" from infinitely far away to inside the sphere at r = R/2

Answer is W = 11/8 (kQ/R) q

Homework Equations



for 0<r<R
E1 = KQR/r^3
Something doesn't look right with the above equation for E1. If you show your work we might be able to help figure out what went wrong. (It looks to me as though you have a couple of variables switched around.)

for r>R
E2 = KQ/R^2
Okay, E2 looks like the same problem. Are you accidentally swapping the r and R?

The Attempt at a Solution



q∫KQR/r^3 + q∫KQ/R^2
Of course you'll need to make corrections to the terms under the integrals, since they're not quite right. Also, they are path integrals, so you'll need to define the paths. (Start at infinity and go to R, then go from R to R/2).

Also, keep in mind that
[tex] W = \int_P \vec F \cdot \vec{dl} [/tex]
is the work done by the sytem, --the work done by the force -- meaning the work is positive if the force and the direction of the path is in the same direction. On the other hand, the problem statement here asks you to determine the work done on the system, by moving the charge: the work done on the force. Because of that, there is a negative sign involved,
[tex] W = -\int_P \vec F \cdot \vec{dl} [/tex]
 
  • #3
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After integration

Kq/R + KQ/R^3 (.4704)

KQ/R ( 1 + (1/R^2) (.4704))

KQ/R (11/8) q = answer
 
  • #4
collinsmark
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After integration

Kq/R + KQ/R^3 (.4704)

KQ/R ( 1 + (1/R^2) (.4704))
Sorry, but I don't quite follow what you're doing there.

You're adding one thing that has R in the denominator to another thing that has R3 in the denominator. The terms don't even have the same dimensions. I'm not sure where the .4704 comes from.

KQ/R (11/8) q = answer
That is the correct answer, yes.
 

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