Applications of Gauss' Law - two hollow spheres

[knowlewj01]

Homework Statement

A hollow sphere of radius r_1 is placed at the centre of a larger hollow sphere of radius r_2.
Both spheres have a uniformly distributed total charge of +q
Find the preassure p(r_2 , q) which acts on the outer sphere.

Homework Equations

\oint\textbf{E.n}dS = 4\pi k Q

p = \frac{dF}{dA}

The Attempt at a Solution

Inside the larger sphere, there is no contribution to electric field from the large sphere, so just consider the electric field at r_2 due to the inner sphere:

E(4\pi r_2^2) = 4\pi k q

E = \frac{kq}{r_2^2}

since F = qE

the force acting at r_2:

F(r_2) = \frac{kq^2}{r_2^2}

p = \frac{F}{A}

p = \frac{kq^2}{4\pi r_2^4}

does this look right?

 

[knowlewj01]

Just had a thought, at the point when calutating force, replace k with \frac{1}{4\pi \epsilon_0}

F(r_2) = \frac{q^2}{4\pi r_2^2 \epsilon_0}

the expression for the total area, A, of the outer sphere is now on the denominator:

A = 4\pi r_2^2

it follows that:

F = \frac{q^2}{A\epsilon_0}

now differentiate F with respect to A:

\frac{dF}{dA} = -\frac{q^2}{A^2 \epsilon_0} = p

p = \frac{q^2}{16\pi^2 r_2^4 \epsilon_0}

gthis is the same answer i got before but i guess it's a more thorough way of doing it.

 

[nikmal]

i took the center sphere to act as a point charge, and calculated E then F then P

got exactly the same. so looks good?