You are here:

Physics/Electric field, potential

Advertisement


Question
plane
plane  
ring
ring  
QUESTION: 1)   Hi, I was just wondering if it would be possible to get some help with these problems. I don't seem to understand what to do at all and these are the only two i have left for my midterm studying which i cannot seem to get. Thanks for any help. A picture is provided to better visualize the problem.

Consider a uniformly-charged ring having a radius of R and a total charge of q_ring.

(a) What is the linear charge density of the ring? Express your answer in terms of R and q_ring
(b) Consider a small element of arc angle delta theta and position theta. What is the charge delta q of this element? Express your answer in terms of R, q_ring, delta theta and theta
(c) What are (i) the electric field delta E, (ii) the electric potential delta V, at P due to delta q? Express your answer in terms of R, q_ring, delta theta, theta and x
(d) What are (i) the net electric field E, (ii) the net electric potential V, at P due to the entire ring? Use symmetry wherever possible, and evaluate your final integrals. Express your answer in terms of R, q_ring and x
(e) Verify that Ex = -dV / dx."

2)    Consider a finite line segment having a linear charge density lamda(x) = 10 |x| (x is in m and lamda is in C/m). The line segment extends to a length of L (not shown) on either side of the origin.

Hint: In setting up the integrals, break the rod into two symmetric parts.


(a) Consider a small element of length delta x and position x. What is the charge delta q of this element? Express your answer in terms of lamda, x and delta x.
(b) What is the total charge of the rod? Express your answer in terms of lamda and L.
(c) What are (i) the electric field delta E, (ii) the electric potential delta V, at P due to delta q? Express your answers in terms of lamda, x, delta x and y.
(d) What are (i) the net electric field E, (ii) the net electric potential V, at P due to the entire rod? Use symmetry wherever possible, and evaluate your final integrals. Express your answers in terms of lamda, y and L.

ANSWER: Hello Timmy,

1a. The circumference is 2*pi*R = so the linear charge density is q_ring / (2*pi*R).

b. The length of the element is R*DeltaTheta, so the charge contained in that element is
(linear charge density) * (length of the element) = [q_ring / (2*pi*R)] * R*DeltaTheta

ci. The distance from DeltaQ to P is sqrt(R^2 + x^2). The field formula is
E = k*q/r^2
Therefore
DeltaE = k*DeltaQ/r^2
where k = 9*10^9 N.m^2 / C^2, q = your result for 1b, and r = sqrt(R^2 + x^2). If you knew the units of q_ring and R, your result would have units of N/C.

cii. The electric potential formula is
DeltaV = k*DeltaQ/r^2
where the parameters are the same as above. This would have units of N.m/C which is equivalent to the volt.

di. Integration will be required. Consider the differential charge de, which is analogous to DeltaQ. The charge dq causes a field dE at P. (dE is analogous to DeltaE.) From symmetry, the total field resulting from integration over the entire ring must lie along the ring axis. So only the component of the field resulting from dq that is parallel to the axis contributes to the total E at P. The component perpendicular to the axis can be ignored.
So E = Int[dE*cos(Psi)]
The function cos(Psi) = x / (R^2 + x^2)
Look at the result for part b which becomes part of the results for the following sections. Convert arclength R*DeltaTheta to DeltaS. So use the analogous differential arclength ds in the integral. Everything (except the ds) in the integral is a constant. So you get those constants times the integral of simply ds. And then the integration just yields the circumference of the ring.

This is still quite a bit of work, so I'll tell you: When you perform the calculation the result should be
E = k*q*x / sqrt(R^2+x^2)

dii. In the same manner, convert DeltaV to dV and integrate.

e. The differentiation isn't going to be easy, but it is basic calculus.

2. The difference between 1 and 2 is a difference between geometry. If you do need further help, I welcome followup questions.

I hope this helps,
Steve

---------- FOLLOW-UP ----------

QUESTION: Thanks for your help

for number 2
I got part a) to be delta q = lambda(delta x)

part b) q_total = integral from 0 to L of 10xdx = 10L^2

but the problem starts at part c and part d. I have no idea how to proceed with those problems.
Also, if the initial function was 10xdx instead of 10|x|dx, what would that change?

Answer
Hello Timmy,

I have minor problems with your results for 2a and 2b. In my experience, some instructors would not accept answers with those minor problems (and some would).

2a. The instructions said to answer in terms of lambda, x, and DeltaX. I suggest that you should answer
DeltaQ = lambda(x)*DeltaX

2b. I agree with your final result, but you didn't explain that you multiplied the integration result (5x^2) by 2 because of symmetry.

2ci.  The distance from DeltaQ to P is sqrt(y^2 + x^2). The field formula is
E = k*q/r^2
Therefore
DeltaE = k*DeltaQ/(y^2 + x^2) = k*lambda(x)*DeltaX/(y^2 + x^2)

2cii. The electric potential formula is
DeltaV = k*DeltaQ/r =
So
DeltaV = k*lambda(x)*DeltaX / sqrt(y^2 + x^2)

2di. Integration will be required. Consider the differential charge dq, which is analogous to DeltaQ. The charge dq causes a field dE at P. (dE is analogous to DeltaE.) From symmetry, the total field at P resulting from integration must be perpendicular to the 2L long line segment. So only the component of the field resulting from dq that is perpendicular to the line segment contributes to the total E at P. The component parallel to the line segment will be cancelled by the field from a symmetrical segment on the other side of center and can be ignored.
So E = Int[dE*sin(theta)]
where theta is the angle between a line from x to P and the line segment.
The function sin(theta) = y/sqrt(y^2 + x^2)
Look at the result for DeltaE in part ci. Use the analogous differential dx in place of DeltaX in the integral.

So
E = Int[(k*10|x|/(y^2 + x^2)) * y/sqrt(y^2 + x^2) dx]  
Looks like a difficult integration. Good luck with that.

2dii. The process here is much like it was for 2di. I'll leave that for you.

If you remove the absolute value bars, the polarity of the charge is negative when x is negative.

I hope this helps,
Steve

Physics

All Answers


Answers by Expert:


Ask Experts

Volunteer


Steve Johnson

Expertise

I would be delighted to help with questions up through the first year of college Physics. Particularly Electricity, Electronics and Newtonian Mechanics (motion, acceleration etc.). I decline questions on relativity and Atomic Physics. I also could discuss the Space Shuttle and space flight in general.

Experience

I have a BS in Physics and an MS in Electrical Engineering. I am retired now. My professional career was in Electrical Engineering with considerable time spent working with accelerometers, gyroscopes and flight dynamics (Physics related topics) while working on the Space Shuttle. I gave formal classroom lessons to technical co-workers periodically over a several year period.

Education/Credentials
BS Physics, North Dakota State University
MS Electrical Engineering, North Dakota State University

©2016 About.com. All rights reserved.