# Calculate the electric potential at the center of the hexagon

1. Apr 7, 2010

### noob314

A plastic rod with a total charge Q uniformly distributed along its length is bent into a regular hexagon where each side has a length of 2a, as shown below. Calculate the electric potential at the center of the hexagon (relative to the point at infinity).

I wasn't sure how to exactly start this problem, so what I did was "simplify" the hexagon into 6 points with a charge of Q/6 on each end of the hexagon and calculated the electric potential of each point from the equation V = kQ/r, giving the total potential of kQ/2a at the center. I'm not sure if that's the correct way to go about in solving this problem, as I don't know the final answer, so any help is appreciated. Thanks.

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2. Apr 7, 2010

### vela

Staff Emeritus
Good try, but no cigar.

By symmetry, you can see that the potential will just be six times the potential due to one segment. To calculate the potential due to one segment, you're going to have to do an integration. Your textbook should have an example of how to do problems requiring integration.

3. Apr 7, 2010

### noob314

Ah, I see. That makes things a lot more simple. So the integral would be
$$V= -\int^{a}_{0}\frac{k\lambda dx}{\sqrt{x^{2}+y^{2}}}$$

with $$\lambda$$ being $$\frac{Q}{6a}$$ and after doing the integral,

$$V = \frac{kQ}{6a}ln\frac{a+\sqrt{a^{2}+y^{2}}}{y}$$

With 6 segments, the total electric potential at the center would be
$$V = \frac{kQ}{a}ln\frac{a+\sqrt{a^{2}+y^{2}}}{y}$$
correct?

4. Apr 7, 2010

### vela

Staff Emeritus
You generally have the right idea, but you need to clean up some details. For instance, your charge density is wrong because the length of each side is 2a, not a. You need to get rid of y and write the answer in terms of a. Also, check the limits on your integral.

5. Apr 7, 2010

### noob314

Edit: I looked at another source for the same example of the rod and found that my textbook's example is incorrect. The limits of integration should be -a to a. Charge density is Q/12a. And the electric potential of one segment would be V= kQ/12a ln[...]

I'm guessing y is a constant, which would therefore make it 2a and the charge density would be Q/12a.

I'm confused about the integral now. From an example I have seen, a rod of finite length L has a charge of q and you have to find the electric potential at a point P at a distance d from the midpoint of the rod.

It looks like they divided the rod into two pieces, and showed through symmetry that those two pieces contribute an equal amount to the overall potential at the point which would give this integral with dq being q/L.
$$V = \int^{L/2}_{0}\frac{2kdq}{r}$$

$$\frac{2kq}{L}ln\frac{a+\sqrt{a^{2}+d^{2}}}{d}$$

Assuming one segment of the hexagon is equivalent to a rod, and substituting my values into their answer, the potential for one segment of the hexagon would be
$$V = \frac{kQ}{6a}ln\frac{a+\sqrt{5a^{2}}}{a^{2}}$$

which is why I'm confused. Since substituting 2a into L in their answer, which would make the 2's cancel out, and substituting Q/6 into q would give me the answer I got originally.

Last edited: Apr 7, 2010
6. Apr 7, 2010

### vela

Staff Emeritus
You're work is good up to here. You can calculate what d is in terms of a using basic trig and geometry.
You replaced q=Q/6 and L=2a, but your expression is still the potential due to just one rod. You need to multiply by 6 (and fix d) to get the potential due to all of the rods.

7. Apr 8, 2010

### noob314

I know that is the potential for one rod, but you stated that I needed to check my limits of integration. I'm also getting conflicting answers from two different sources. In one source, they integrate from -a to a, and have the electric potential as

$$\frac{kQ}{L}ln\frac{a+\sqrt{a^{2}+d^{2}}}{-a+\sqrt{a^{2}+d^{2}}}$$
and in another source, they integrate from 0 to a and their answer is

$$\frac{2kQ}{L}ln\frac{a+\sqrt{a^{2}+d^{2}}}{d}$$
I'm not sure which one is correct or if those two are equivalent to each other.

8. Apr 8, 2010

### vela

Staff Emeritus
Oh, okay. I thought you were saying that was your final answer. Regarding the limits of integration, in your first post, you integrated from only 0 to a yet had no factor of 2. That's what I was getting at.
They're equivalent to each other because

$$\frac{d}{-a+\sqrt{a^2+d^2}}\times\frac{a+\sqrt{a^2+d^2}}{a+\sqrt{a^2+d^2}} = \frac{a+\sqrt{a^2+d^2}}{d}$$

9. Apr 8, 2010

### noob314

Got it. Thanks for helping me out.