Electric Potential and Kinematics

[Elysium]

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

Ok first the problem:

An insulated sphere of radius R is charged unformly with the total charge +Q. Since the electric field is outwards, an electron of charge -e (insulated, so that it won't disappear while traveling iside the charged sphere) placed at the surface will be pull toward the center of the sphere.

• What is the initial potential energy of the electron (relative to a point at infinity in terms of Q and R)?
• Calculate the velocity of the electron (also in terms of Q and R) when it reaches the center of the sphere. Note that the electric field and hence the acceleration is not constant. Use the concept of conservation of kinetic and potential energies. Some simple integration is necessary to compute the potential at the center of the sphere.

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?

 

[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.

 

[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.

 

[durt]

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

 

[Elysium]

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

 

[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).