Electric Field and the Speed of a Proton

[Ithryndil]

Homework Statement

A particle with a charge of -60.0 nC is placed at the center of a nonconducting spherical shell of inner radius 20.0 cm and outer radius 34.0 cm. The spherical shell carries charge with a uniform density of -2.26 µC/m3. A proton moves in a circular orbit just outside the spherical shell. Calculate the speed of the proton.

Homework Equations

We will need:
$$E = F/q_{o}$$
$$\Phi=EA=q_{inside}/\epsilon_{o}$$ (no integral is needed because we know the electric field will be constant at the surface of the sphere and we know the surface area of a sphere).
$$F = ma_{c}$$

The Attempt at a Solution

Solving for E I get:

E = $$q_{inside}/(\epsilon_{o}A)$$

$$q_{o}$$ is just the inner charge (-60.0nC) + the outer charge [4/3*pi*charge density*(0.34^3-0.20^3).

Plugging in for E I get:

$$F/q_{o}=q_{inside}/(\epsilon_{o}A)$$

$$q_{o} = q_{inside}$$ because the spherical surface should act as a point charge right?

Therefore after some algebra and substitution for the centripetal acceleration I get:

$$v = \sqrt{q^{2}/(4\pi\epsilon_{0}rm})$$
Where r = .34 and m is the mass of a proton.

When I plug in all the values I get a speed on the order of $$10^{12}m/s$$
Which is faster than the speed of light if I am not mistaking...that being roughly $$3 x 10^{8}m/s$$

What am I doing wrong?

 

[Ithryndil]

Homework Statement

$$F/q_{o}=q_{inside}/(\epsilon_{o}A)$$

$$q_{o} = q_{inside}$$ because the spherical surface should act as a point charge right?

My problem lies with the above. $$q_{o} = q_{inside}$$. That is a false statement. The $$q_{o}$$ is actually the charge of the proton, not the charge of the entire charge configuration. With that adjustment I get an answer on the order of 10^5 which is must more realistic and was the correct answer.

 

[Defennder]

Yup that's right. Good that you figured it out by yourself.