Review of Electrical Field Calculations

1. Feb 27, 2009

valarking

Hello, I have an assignment in my Electricity and Magnetism course that covers the last few chapters dealing with calculating charge, using Gauss's Law, and electric potential. I'm going through them right now, but physics isn't completely my strong point so I would really appreciate it if I could get an idea of whether I'm approaching the problems correctly.

I'll just go one problem at a time (there are 5).

Problem 1:

The problem statement, all variables and given/known data
A solid insulating finite-thickness spherical shell of the inner radius a and outer radius b
is uniformly charged and carries the total charge Q. Find the magnitude of the electric
field E everywhere in space, including points outside and inside the spherical shell. What
is the electrostatic potential of the outer surface of the shell? If, in addition, you can find
the potential of the inner surface of the shell, you would get bonus points on this problem!

The attempt at a solution
Gauss's Law states that $$\oint {\vec{E} \cdot d\vec{A}} = \frac{Q_{enclosed}}{\epsilon_0}$$.
So since we're talking about a shell with uniformly distributed charge, we can say: $$E = \frac{Q_e}{4{\pi}r^2\epsilon_0}$$, or $$E=k_e\frac{Q_e}{r^2}$$.

Now, assume that I want to obtain the magnitude of the electric field E at points A, B, and C, where A is outside of the hollow sphere, B is in between the inner and outer radii, and C is inside the cavity.

For point A:
All of charge Q is enclosed in the spherical gaussian surface at point A, so $$Q_e = Q$$. So if we take r to be the distance to point A, the field E at point A is calculated by: $$E_{point A} = k_e\frac{Q}{r^2}$$

For point B:
Enclosed charge is ratio of charge up to B over total charge in shell, which is:
$$\frac{\frac{4}{3}(r^3-a^3)}{\frac{4}{3}(b^3-a^3)}$$
So if we take r to be the distance to point b, Q enclosed in this case would be:
$$Q_e = Q(\frac{r^3-a^3}{b^3-a^3})$$
so E would be:
$$E_{point B} = k_e\frac{Q(\frac{r^3-a^3}{b^3-a^3})}{r^2}$$

For point C:
Since there is nothing inside the spherical cavity, and all charge Q is outside of it, that means that there is no charge inside a Gaussian surface centered within the sphere containing point C. Since there is no charge, putting 0 into Q for the electrical field means that:
$$E_{point C} = 0$$

Surface potential:
Since electrical field of a sphere is calculated the same as a point once outside of the sphere, our hollow sphere with charge Q can be considered a point, so the electric potential on the surface can be calculated as $$\phi = k_e\frac{Q}{b}$$

Ok, that's what I got for the first problem.
I'm fairly certain A and C are correct, but I'm wondering if what I did at point B to calculate enclosed charge is correct.

Problem 2:

The problem statement, all variables and given/known data
Figure 1 displays two linear objects aligned along the same axis. One object is a semiinfinite
insulating line charged with the uniform linear charge density $$\lambda$$. The other is
the insulating uniformly charged thin rod. Rod’s total charge is Q and length 2a. The
distance from rod’s center to the end of the semi-infinite line is r. What is the force F
exerted on the rod?

http://www.vkgfx.com/physics/fig1.jpg [Broken]

The attempt at a solution
Electric force is qE, or $$F = k_e\frac{{q_1}{q_2}}{r^2}$$
So in this problem I define those as:
$$q_1 = \lambda{dx}$$
$$q_2 = \frac{Q}{2a}$$
$$r^2 = (r - a + x + y)^2$$
Essentially I'm splitting the infinite line into infinite parts of density $$\lambda$$ and splitting the line of length 2a into infinitely small pieces of density $$\frac{Q}{2a}$$. I will integrate them with x being along the infinite line and y being along the 2a line. Therefore the distance between each infinitely small piece will be the distance from the end of the infinite line to the 2a line (r-a) plus x and y, the distances along each line.
So here's my integral:
$${k_e}\int_0^{2a}\int_0}^{\infty}\frac{\lambda\frac{Q}{2a}}{(r-a+y+x)^2}dxdy$$

I'll skip over most of the grunt calculus work, but the first integration yields:
$$\frac{k_e\lambda{Q}}{2a}\int_0^{2a}{\frac{dy}{r-a+y}}$$
Integrating this gives:
$$\frac{k_e\lambda{Q}}{2a}ln\left|\frac{r+a}{r-a}\right|$$

I believe that's the answer. That's a fairly in depth problem though, so I could be off. Would appreciate any input.

Problem 3:

The problem statement, all variables and given/known data

Figure 2 shows two arrangements of “infinite” parallel plate objects. In both cases,
there are two thin plates carrying total surface charge densities $$\sigma$$ and $$\sigma_1$$ respectively
(polarities of those charges are not specified and should be treated algebraically). An
uncharged conducting slab shown as the filled shape is placed in between the charged
plates in configuration (a) but outside the charged plates in configuration (b). What
are the induced charged densities on both surfaces of the conducting slab in each of the
configurations?

http://www.vkgfx.com/physics/fig2.jpg [Broken]

The attempt at a solution

Note, I refer to $$\sigma$$ in the problem as $$\sigma_0$$ for clarity.
For this problem I put:
a.
top of slab: $$\sigma_0-\sigma_1$$
bottom of slab: $$\sigma_1-\sigma_0$$
b.
top of slab: $$-(\sigma_0+\sigma_1)$$
bottom of slab: $$\sigma_0+\sigma_1$$

My work is mainly visual and right now I'm working to expand on it to explain those symbolically.

Problem 4:

The problem statement, all variables and given/known data

A positively charged bead of charge q and mass m can move without friction along a
straight rod, as sketched in Figure 3, in the presence of another fixed-position charge
−5q. The bead is observed to execute an oscillating motion between points A and B
symmetrically positioned around the fixed-position charge and spaced by the distance
2a. What is the velocity v of the bead when it crosses point C exactly “against” the
fixed-position charge? The distance from the fixed charge to the rod is b.

http://www.vkgfx.com/physics/fig3.jpg [Broken]

The attempt at a solution

I used conservation of energy.
$$K_i + U_i = K_f + U_f$$

In this case I considered initial to mean the bead is at point A, and final to mean it is passing point C.

$$K_i = 0$$
$$U_i = \frac{-5q^2}{\sqrt{a^2+b^2}}$$
$$K_f = \frac{1}{2}mv^2$$
$$U_f = \frac{-5q^2}{b}$$

Giving me:
$$\frac{-5q^2}{\sqrt{a^2+b^2}} = \frac{1}{2}mv^2 - \frac{5q^2}{b}$$
or
$$v^2 = \frac{10q^2{k_e}\left(\frac{1}{b} - \frac{1}{\sqrt{a^2+b^2}}\right)}{m}$$
which means solving for v yields:

$$v = \sqrt{\frac{10q^2{k_e}\left(\frac{1}{b} - \frac{1}{\sqrt{a^2+b^2}}\right)}{m}}$$

Problem 5:

The problem statement, all variables and given/known data

Three conducting spheres of radii a, b and c, respectively, are connected by negligibly
thin conducting wires as shown in figure 4. Distances between spheres are much larger
than their sizes. The electric field on the surface of the sphere of radius a is measured to
be equal to $$E_a$$. What is the total charge Q that this system of three spheres holds? How
much work do we have to do to bring a very small charge q from infinity to the sphere of

http://www.vkgfx.com/physics/fig4.jpg [Broken]

The attempt at a solution

This one I wasn't so sure about. In order to find the total charge, I first applied Gauss's Law to find the charge of one sphere, A, given the electric field on its surface. So $$\oint {\vec{E} \cdot d\vec{A}} = \frac{Q_{enclosed}}{\epsilon_0}$$ is used to get:
$$4\pi{a^2}E = \frac{q_a}{\epsilon_0}$$, or $$q_a = 4\pi{a^2}\epsilon_0{E}$$.

To get the relationship between one sphere and the total charge in a system of conducting connected spheres, I used this equation, which I'm not so sure about.
$$q_A = \frac{Qa}{a+b+c}$$
I'm not sure if the radii should be squared.

Solving for Q gives:
$$Q = 4\pi\epsilon_0{a}E(a+b+c)$$

Now for the second part, again I wasn't sure of this. But you could say the work going from infinitely far to right at sphere B would be the difference in potential energies.
So:
$$k_e\frac{q_b}{b} - k_e\frac{q_b}{\infty}$$
where the infinity cancels out the second fraction and leaves:
$$k_e\frac{q_b}{b}$$