Physics Questions - Electric Fields, Electric Potential, etc

1. Jan 26, 2005

alexman

Thank you for taking the time to review my work. I am posting problems with solutions, but I'm not sure if my solutions are correct. The problems are exactly how I have them written in front of me. These are mostly conceptual questions, which require a short answer of some sort.

Q1. A charge Q coulombsis placed at the origin of co-ordinates. Give the three vector force componentson a charge -Q at a point P(x,y,z).

My Answer: Fx = (k|Q||-Q|)/(r^2); Fy = (k|Q||-Q|)/(r^2); Fz = (k|Q||-Q|)/(r^2). The three vector forced components on charge -Q (at point P) would be on each of the three different axes: x,y, and z. After some calculations, you can use all three to find the direction and magnitude of charge -Q at point P.

Q2. A charge q is placed fixed at the origin. A thin hollow metal sphere radius a is centered at the origin and a charge of q is placed initially on the outside surface. After the charge has rearranged itself give the electric field
(i) at a point P(x,y,z) inside the sphere, r<a;
(ii) in the medal of the sphere, r=a;
(iii) outside the sphere, r>a.
(iv) What is the charge on the inner metal surface?

(i) The electric field must be zero because it is that equilibrium inside the sphere. E=0
(ii) the electric field in the metal of the sphere is equal to charge q at the origin.
(iii) Outside this year, the electric field is E=(q)/(4)(pi)(Eo)(r^2). the electric field outside is uniformly spread along the surface. [got equations from book, not sure if correct].
(iv) the charge on the inner metal surface would be opposite or negative to the charge q.

Q4. Give the potential as a function of r for the three regions in problem Q2. Hint: use the fact that electric fields and potentials from different charges add. And the standard results that a sphere of radius a centered at the origin and uniformly charged with q has potential kq/a if r<a and kq/r if r>a.

Using the hint, this is what I came up with...It's not much, but I hoping someone could help me out some more.

From Q2, using the info from the problem itself:
(i) kq/a
(ii) a=r so then kq/a = kq/r
(iii) kq/r

kq/a + r + kq/r

get them all the same denominator

kqr/ar + ar^2/ar + kqa/ar

so then i came up with the equation which leads me no where really

(kqr + ar^2 + kqa)/(ar) <-----this would then equal the potential as a function of r....somehow

Hopefully you have better luck with any of these problems than I did. Any input on any question would be great. Thanks for you time.

alexman

2. Jan 26, 2005

alexman

this homework is due tomorrow, so please at least try and review some of my work

thanks

3. Jan 26, 2005

learningphysics

Fx, Fy and Fz are wrong. You want the x, y and z components of the force vector.

No. Remember it is a hollow sphere. This part is inside the hollow area. Use gauss law here to find the electric field.

Here your previous answer of E=0 makes sense... as this is inside the metal of sphere.

Yes this is right. Just using gauss' law, gives the answer here.

Yes this is correct. This can be seen using gauss law, on a gaussian sphere whose radius is between the inner and outer radius of the sphere.

This sphere problem is kind of confusing as they have not mentioned the inner and outer radius... they just make you accept that r=a is the metal of the sphere... Your results aren't right. I suggest drawing a diagram... there should be 3 sets of charges. The first is the point charge at the center, the second is the charge of -q distributed on the inner surface, and the 3rd is the charge distribution of +q on the outer surface. Do you see how you get these 3 charge distributions?

So you have a point charge in the center... what is the potential due to the point charge in region I, II, III.

Find the potential due to the inner surface charge distribution in all three regions I, II, III.

Find the potential due to the outer charge distribution in all three regions I, II, III.

Finally add the different potentials in region I to get total potential in region I. Do the same for regions II and III. Be careful with signs.

For this potential problem, I suggest using inner radius=a1, outer radius = a2... solve the potential problem, then finally set a1=a2=a.

So the 3 regions are r<a1 a1<r<a2 r>a2.

Last edited: Jan 26, 2005
4. Jan 26, 2005

alexman

thanks learningphysics, i'll get right on it

I'll post my new results. Thanks for taking the time to review my problem, I really do appreciate all your help.