# Physics problem with electrodynamics

1. Nov 8, 2009

### kliker

hello dear friends, i have another problem in physics, it has to do with electrodynamics.

if we have a sphere with total loading Q which is uniformly distributed throughout it's volume, how can we find the voltage for R/2<=r<=R and for r>=R? (assume that we know E).

i know that Va - Vb = integral(ra/rb) of Edl

but I cant understand how to use it here

2. Nov 8, 2009

### lanedance

this is a good candidate for guass's law

3. Nov 8, 2009

### kliker

it would be a good candidate (i guess) if we didnt know the value of E, but as i said we know E for (1) R/2<=r<=R and (2)r>=R

(1) E = k*Q*r/R^3
(2) E = k*Q/r^2

4. Nov 8, 2009

### kliker

for r>=R/2

if we say V(inf) = 0

then for r > R/2

we will have

V(r) - V(inf) = Integral[r,inf] * Edl = (1/4p*eo)*Q/r

for r = R

we have V(R) = (1/4p*eo)*Q/R

so for r>=R/2

it will be V(r) = (1/4pe) * Q* (1/r - 1/R) ?

:S

5. Nov 8, 2009

### lanedance

you method looks reasonable, though i'm not so sure about the 2nd integration part...
so following what you have done, you mean for r>R and E = k*Q/r^2
$$V(r) = V(r) - V(\infty) = - \int_{\infty}^r \vec{E} \bullet \vec{dr} = - \int_{\infty}^r dr \frac{kQ}{r^2} = \frac{kQ}{r}$$

note as the field is only dependent on r, we can differentiate reasonably easy & it gives the correct field, and the potential goes to zero as required, so we're looking good
$$\vec{E} = -\nabla V(r)= -\frac{dV(r)}{dr}\vec{\hat{r}} = \frac{kQ}{r}\vec{\hat{r}}$$

and fidning the correct offset potential for the edge of the sphere is
$$V(R)= \frac{kQ}{R}$$

then for r<R, inside the sphere, integerate similar to before & carry the constant for V(R) so the potential is at the correct offset
$$V(r)-V(R) = - \int_{R}^r \vec{E} \bullet \vec{dr}$$

now if you differentiate the potential you have found, you don't get the correct field, so i would look have a look at the integral again
$$\vec{E} = -\nabla V(r)= -\frac{d}{dr}( \frac{kQ}{r}-\frac{kQ}{R})\vec{\hat{r}}= \frac{kQ}{r^2}\vec{\hat{r}}$$ ???