Work on charge from outside to inside of sphere

[origamipro]

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

 

[collinsmark]

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
$$W = \int_P \vec F \cdot \vec{dl}$$
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,
$$W = -\int_P \vec F \cdot \vec{dl}$$

 

[origamipro]

After integration

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

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

KQ/R (11/8) q = answer

 

[collinsmark]

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.

 

[Nickie]

a) For 0<r<R, the electric field can be found using the equation E = KQr/r^3, where K is the Coulomb's constant, Q is the total charge of the sphere, and r is the distance from the center of the sphere. In this case, r can range from 0 to R. So, the equation becomes E = KQ/R^2.

For r>R, the electric field can be found using the equation E = KQ/R^2, where R is the radius of the sphere. In this case, r can range from R to infinity. So, the equation becomes E = KQ/r^2.

b) To bring a point charge "q" from infinitely far away to inside the sphere at r = R/2, work needs to be done against the electric field of the sphere. The work done can be calculated using the equation W = q∫E.dr, where E is the electric field and dr is the infinitesimal displacement.

For r>R, the work done is given by W = q∫KQ/R^2 dr = q(KQ/R)∫dr = q(KQ/R)(r-R), where r ranges from R to R/2. So, the work done is W = q(KQ/R)(R/2-R) = q(KQ/R)(R/2-R) = 11/8 (kQ/R) q.