Calculating Electric Field with Point Charge and Hollow Sphere

[jimbo007]

if u have an uncharged,perfectly conducting hollow sphere of thickness t and radius R, and a point charge Q is placed a distance D from the centre of the sphere so that R-t>D, what would the electric field, E, be?
i just treated it the same as if Q was in the centre of the sphere and the electric field lines came perpendicular to the sphere and so

E=\frac{Q}{4 \pi \epsilon_{0}D^2}

anyone got any better ideas?

 

[pmb_phy]

if u have an uncharged,perfectly conducting hollow sphere of thickness t and radius R, and a point charge Q is placed a distance D from the centre of the sphere so that R-t>D, what would the electric field, E, be?
i just treated it the same as if Q was in the centre of the sphere and the electric field lines came perpendicular to the sphere and so

E=\frac{Q}{4 \pi \epsilon_{0}D^2}

anyone got any better ideas?

There is no reason to assume that you can do that. One the charge is offset from the center you've changed he symmetry of the situation.

Pete

 

[Galileo]

The point charge induces a charge (-Q) on the inner surface as to cancel the field inside the conductor. The remaining induced charge (Q) will distribute itself over the surface of the sphere.
I'm pretty certain it will spread uniformly so that the electric field you got is correct.

 

[Doc Al]

E=\frac{Q}{4 \pi \epsilon_{0}D^2}

I assume you meant to write:
E=\frac{Q}{4 \pi \epsilon_{0}r^2}
That would be correct for the field for r > R. (r = 0 is the center of the sphere.)

See my answer here: https://www.physicsforums.com/showthread.php?t=37090

 

[turin]

There is no electric field inside the conductor. The conductor accomplishes this by arranging a layer of charge in an appropriate distribution on the outer surface of the sphere with a total amount equal to -Q. The inner surface of the sphere therefore develops a negative charge deficit equal to this (+Q) and also distributed appropriately (since there is no field inside the conductor). This all results in a negative image charge somwhere below the surface of the sphere, not quite in the center, and not necessarily near the surface, and not necessarily equal to -Q. I don't remember what the expression is off the top of my head; sorry. Look up the "method of images."

 

[jimbo007]

lol, thanks for the responses guys, very much appreciated
but I'm still completely stumped by the question
just to make it clear i want to find an expression for the electric field outside the sphere (after reading my first post i didnt think my question was very clear)

thanks again

 

[Doc Al]

lol, thanks for the responses guys, very much appreciated
but I'm still completely stumped by the question
just to make it clear i want to find an expression for the electric field outside the sphere (after reading my first post i didnt think my question was very clear)

What's stumping you? I gave the expression for the field outside the sphere in my first response. Now... to explain why that's the right answer requires a little more...

 

[Doc Al]

There is no electric field inside the conductor.

Right.

The conductor accomplishes this by arranging a layer of charge in an appropriate distribution on the outer surface of the sphere with a total amount equal to -Q.

I believe you have it backwards. The -Q is induced on the inner surface. This surface charge is arranged to exactly cancel the field from the charge within the cavity, thus producing zero field everywhere inside the conductor.

 

[jimbo007]

i just typed out my response and realized something. i couldn't see the difference between the formula i gave and the formula u gave but now i think i see it.
so by the sounds of it, it doesn't matter where you place the charge Q inside the sphere, E outside the sphere depends on the position only from the centre of the sphere. is this right?

 

[Doc Al]

That is correct.

 

[turin]

Doc Al,
Oops. I read the post too quickly. I was responding under the false impression that the point charge, Q, was placed outside the conductor. Sorry to all.