Basic Electric Field and Symmetry Problem - help

[fatcat39]

Homework Statement

24 identical positive charges, each has a magnitude 2.0 microCoulombs equally spaced along a circle with a radius of 0.4 m which lies in the y-z plane.

a)determine the electric field at a point 1 meter away from the circle's center along the x-axis.

b) Find the electric potential at the point P

c) What is the electric field and potential at the circle's center?

Homework Equations

E = kQ/R^2
V = kQ/R

The Attempt at a Solution

I know that symmetry applies, but I'm not sure exactly how.

A) 2 sigfigs
Positive - radiate outwards to point P
E = kQ/R^2
- Should I calculate the charge for each individual charge for 12 of them and multiply that answer by 2?
- Or can I calculate the field for one and multiply by 24?

C) The electric field at the center of the circle is 0, I think, because the charges cancel each other out.

D) But does that mean that the potential at the center is 0 as well?

 

[Rainbow Child]

1) Since you have 24 identical positive charges, this means that you actually have 12 pairs, each pair lying at the end of a diameter. Can you calculate the electric field of eah pair. Hint: You have to and vectors.

2) For the circle's center. The potential is a vector or a scalar?

 

[fatcat39]

1) So with each pair: the y-components cancel each other out. But the x-components are doubled, yes? And then the electric field of that is multiplied by 12?

2) The potential is a scalar, so...hmm..

 

[Rainbow Child]

1) So with each pair: the y-components cancel each other out. But the x-components are doubled, yes? And then the electric field of that is multiplied by 12?

Correct!

2) The potential is a scalar, so...hmm..

...so find the potential of one of the charges and multiply by 24.

 

[fatcat39]

Okay, so ...

Pythagorean theorem
.4^2 + 1^2 = 1.16
$$\sqrt{1.16}$$


A) E = 2 * kQ/R$$^{2}$$ * cos $$\vartheta$$
12 pairs --> 12 * 11525.91754 = 138311.0105
E = 1.4 * 10$$^{5}$$ N/C

B) V = kQ/R
24 V = 24 * 9 * 10$$^{9}$$ * 2 * 10[tex]^{-6} / $$\sqrt{1.16}$$ = 401101.93<br /> = 4 * 10^{5} V<br /> <br /> C) Electric field at circle's center = 0 = Symmetry<br /> <br /> D) Electric potential at circle's center = <br /> <br /> 24V = 24 * 9 * 10^9 * 2 * 10^-6 / .4 = 1080000 = 1.1 * 10^6<br /> <br /> Can you check these please? Thanks so much![/tex]

 

[Rainbow Child]

A) E = 2 * kQ/R * cos
12 pairs --> 12 * 11525.91754 = 138311.0105
E = 1.4 * 10 N/C

I am not sure that's correct, but my calculations are terrible!

 

[fatcat39]

I think the cosine part might be off...Hmm.

 

[Rainbow Child]

$$\cos\theta=\frac{1}{\sqrt{1.16}}$$, yes?

 

[fatcat39]

Why isn't it cos theta = .4/sqrt 1.16?

 

[Rainbow Child]

Because the angle $\theta$ is between the axis x and the line joining point P and a charge. Isn't it?

 

[fatcat39]

Isn't it between the z-axis and the line between P and a charge? I'm drawing that out of an example in my textbook, but the sample charges it showed were + 1 and - 1, not 2 positive charges. Would that make a difference?

 

[Rainbow Child]

You said that point P is

1 meter away from the circle's center along the x-axis

That's why I said x-axis.

 

[fatcat39]

I guess I'm a little confused about where the triangle is, because the picture I have, z is vertical, x is horizontal, and y is halfway between them - so I thought that the triangle has legs of z for the base and 1.0 m as the height for P. But still, isn't theta (if you visualize the isoceles triangle) one of the two identical base angles?

Thanks so much for helping me.

 

[Rainbow Child]

Ok, orget the axis!
Let's call the center C, point P is point P and the charge under investigation is point Q and let's call $\theta$ the angle formed by CP and PQ.

The electric field is along QP and makes with CP the same angle $\theta$.
The compontents that survive are along CP, and the cos is given by

$$\cos\theta=\frac{CP}{PQ}=\frac{1}{\sqrt{1.16}}$$

Is it clear now?

 

[fatcat39]

Ohhhhhhhhhhhhhh...okay. :) Thanks a lot!

One more quick question:
"A thin ring of the same radius has a uniformly distributed charge of + 1 microCoulombs. What is the E field 2.0 m away along the x axis?"
N approaches infinity = number of points to form the ring
E = (N/2) (kQ/R^2) cosine theta

Does that work? And if it does, what do I substitute for N if N has to approach infinity?

 

[Rainbow Child]

Since we have a uniformly distributed charge, call it $Q$ we split it in N point charges $q=\frac{Q}{N}$.
Then for one pair we have

$$E=2\,k\,\frac{q}{(QP)^2}\,\cos\theta \RightarrowE=2\,k\,\frac{Q}{N\,(QP)^2}\,\cos\theta$$

Thus the total electric field

$$E_t=\frac{N}{2}\,E\Rightarrow E_t=k\,\frac{Q}{(QP)^2}\,\cos\theta$$

 

[fatcat39]

Theta being 2/ sqrt(4.16), right?

(LaTex doesn't work so well on my computer)