Review of Electrical Field Calculations

[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:

Homework Statement
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:

Homework Statement
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

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:

Homework Statement

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

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:

Homework Statement

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

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:

Homework Statement

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
radius b?

http://www.vkgfx.com/physics/fig4.jpg

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:
$$k_e\frac{q_b}{b}$$
as the answer.

I believe that any effect a or c would have on charge q's potential would be small enough not to matter. I am correct in this assumption?

Edit:
Ok, I finished it and have my attempts posted. I would really appreciate any input that I could get about my methods, as I am not too comfortable with this subject.
Thanks!

 

[anathema]

Thank you for sharing your attempts at solving these problems. it is always great to see students actively engaging with their coursework and seeking help when needed.

Overall, your approach to these problems seems to be on the right track. However, there are a few areas where some clarifications or corrections may be necessary.

In problem 1, you correctly use Gauss's Law to find the electric field outside and inside the charged spherical shell. However, your calculation for the electric potential on the outer surface of the shell is not entirely correct. The correct formula for the electric potential on the surface of a charged spherical shell is \phi = k_e\frac{Q}{r}, where r is the distance from the center of the shell to the point of interest. This gives you the potential on the outer surface as \phi = k_e\frac{Q}{b}. Similarly, the potential on the inner surface can be found using the same formula, but with r = a, giving \phi = k_e\frac{Q}{a}.

In problem 2, you are on the right track by using the formula for electric force, but your integration is not correct. To find the force on the charged rod, you need to integrate the force between each infinitesimal charge element on the rod and each infinitesimal charge element on the semi-infinite line. This would give you an integral of the form \int_{0}^{2a} \int_{0}^{\infty} \frac{\lambda dx dy}{(r-a+y+x)^2}, where x and y are the distances along the rod and the semi-infinite line, respectively. This integral can be simplified using trigonometric identities and the final result would be F = \frac{k_e\lambda Q}{r^2}\left(\frac{1}{r-a}-\frac{1}{r+a}\right).

In problem 3, your approach seems to be correct, but I would suggest labeling the surfaces of the conducting slab as top and bottom instead of using the terms "above" and "below". This will make it clearer which surface you are referring to.

In problem 4, your application of conservation of energy seems to be correct. However, the electric potential energy should be calculated for the entire system, not just the fixed charge and the bead. This would give U_i = \frac{-5q^2}{\sqrt

 

[nlightner]

Dear student,

Thank you for sharing your review of electrical field calculations. Your approach to the problems seems to be correct and you have applied the relevant equations and concepts appropriately. However, I would like to provide some feedback on a few points:

1. In problem 1, your approach to calculating the electric field at points A, B, and C is correct. Your calculation of the surface potential is also correct.

2. In problem 2, your approach to finding the force on the rod is correct. However, your integral for the electric field might need some clarification. The distance between the two charged objects should be taken as the distance between the center of the semi-infinite line and the center of the rod, which is r + a. Also, the limits of integration should be from 0 to infinity for both x and y. This will give you the correct expression for the electric field.

3. In problem 3, your approach to finding the induced charge densities on the conducting slab is correct. However, I would suggest using a different notation for the induced charge densities to avoid confusion with the given surface charge densities. For example, you could use \sigma_+ for the induced positive charge density and \sigma_- for the induced negative charge density.

4. In problem 4, your approach to finding the velocity of the bead is correct. However, I would suggest using the conservation of energy equation in terms of potential energy instead of electric field. This will give you a simpler expression for the velocity.

5. In problem 5, your approach to finding the total charge of the system is correct. However, for finding the work done in bringing a small charge q from infinity to the sphere of radius b, you should use the total potential energy equation instead of just the potential energy at point B. This will give you a more accurate result.

Overall, your approach to the problems seems to be correct and you have a good understanding of the concepts involved. Keep up the good work and don't hesitate to ask for help if you have any further doubts. Good luck with your assignment!