# Electric Potential and Kinematics

1. Apr 22, 2006

### Elysium

Hi, I'm currently doing a practice problem and I need help in solving it.

Ok first the problem:

a)Ok, the potential energy of the electron is:

$$U = -eV_R$$

To find V of R we need to refer to electric field:

$$V_f - V_i = -\int_{i}^{f} \vec{E} \cdot d\vec{R}$$
$$V(R) - V(\infty) = -\frac{Q}{4\pi\varepsilon_\varnothing} \int_{R}^{\infty} \frac{dR}{R}$$
...
$$V_R = \frac{1}{4\pi\varepsilon_\varnothing}\frac{Q}{R}$$

Therefore,

$$U = -eV_R$$
$$U = \frac{1}{4\pi\varepsilon_\varnothing}\frac{-eQ}{R}$$

B)This is the one I'm having trouble with. I don't know what the final potential energy is or out to get it.

Here's what I have so far:

$$U_\varnothing = K_f + U_f$$
$$\frac{1}{4\pi\varepsilon_\varnothing}\frac{-eQ}{R} = \frac{1}{2}mv^2 + U_f$$

So what do I do for U of f?

Last edited: Apr 22, 2006
2. Apr 22, 2006

### durt

You'll first have to find the electric field inside the sphere using Gauss's law. Then integrate that from 0 to R to get the potential at the center, designating the potential at the surface zero.

3. Apr 22, 2006

### Elysium

That means:

$$\int \vec{E} \cdot d\vec{S} = \frac{Q_{enclosed}}{\varepsilon_\varnothing}$$
$$\int E \cdot dS cos(0) = \frac{Q_{enclosed}}{\varepsilon_\varnothing}$$
$$E \cdot 4 \pi R^2 = \frac{Q_{enclosed}}{\varepsilon_\varnothing}$$

Charge is intact, so...
$$\int E = \frac{1}{4\pi\varepsilon_\varnothing} \frac{|Q|}{R^2}$$

Which is the same for pointlike charges...

Now:

$$V_f - V_i = -\int_{i}^{f} \vec{E} \cdot d\vec{R}$$
$$V(R) - V(0) = -\int_{0}^{R} \vec{E} \cdot d\vec{R}$$
$$V(R) - V(0) = -\frac{Q}{4\pi\varepsilon_\varnothing} \int_{0}^{R} \frac{dR}{R}$$

eh there's a division by zero. Does this mean that there's no potential energy in the center? That's what I assumed at the begining, but now I'm lost.

Last edited: Apr 23, 2006
4. Apr 23, 2006

### durt

I think you've forgotten Gauss' Law:
$$\int \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon_0}$$

5. Apr 23, 2006

### Elysium

No I haven't. it's in my previous post. Am I complicating things?

6. Apr 23, 2006

### durt

You're trying to find the electric field at any distance $$r$$ from the center of the sphere. So:
$$E \cdot 4 \pi r^2 = \frac{\rho V}{\epsilon_0}$$
where
$$\rho = \frac{Q}{\frac{4}{3} \pi R^3}$$
and
$$V = \frac{4}{3} \pi r^3$$
(Notice r is not the same as R).

Last edited: Apr 23, 2006