# Electric potential at center of semicircle

## Homework Statement

A total charge Q is uniformly distributed on a thread of length L. This thread forms a semicircle. what is the potential at the center? (assume V=0 at large distances)

V = -∫Edl
E = kQ/r^2

## The Attempt at a Solution

V = - ∫Edr = -E∫dr = -Er = - kQr/r^2 = -kQ/r
is this correct? I am not sure if the sign should be negative

mfb
Mentor
E is a vector and it is not constant. Finding the electric field is more complicated, it is sufficient to calculate the potential here but that has a different formula.
Also, using r both for radius and for the length along the circle is confusing and leads to wrong results.

hmmm so would it be correct to say
V = ∫dV = k∫dQ/r = kQ/r because Q is constant?

mfb
Mentor
The result is right but you'll have to justify the steps you used.

The result is right but you'll have to justify the steps you used.
Does this mean i have to plug something in for dQ?

mfb
Mentor
The length should appear somewhere.

Oh, and I guess the final result should be expressed in terms of L instead of r as well I guess.

I think i can use arc length formula L = theta r to relate r and L. I am not sure how I can plug that in to the equation for V. I think this doesn't seem like it should be a tough problem but I must be missing something obvious =[

mfb
Mentor
This is not a tough problem. Yes you can use the arc length formula.

If a charge Q is uniformly distributed over a length L, what is the charge per length?

I think it would simply be Q/L so dQ in terms of the charge density would be dQ = λdL = (Q/L)dL

V = k∫dQ/r = k∫(Q/L)dL/r = kQL/Lr = kQ/r ??

mfb
Mentor
Looks fine. Now you just have to express r in terms of L.
The semicircle has a length of L.

so since L = theta r then r = L/theta so the final step would be
kQ/r = kQtheta/L
but since they dont give me theta can I get away with doing this?

Nathanael
Homework Helper
they dont give me theta

assuming a semicircle is a half circle (forgive me if this is incorrect I am not 100 percent sure) then L would equal the perimeter or L = (pi/2)r^2
so r = sqrt(2L/pi) so it would be kQ/(sqrt(2L/pi))

Nathanael
Homework Helper
assuming a semicircle is a half circle (forgive me if this is incorrect I am not 100 percent sure)
Yes, semicircle just means half-circle

L = (pi/2)r^2
I think you're mixing up the area and circumference formulas. On the left side you have units of length but on the right side you have units of length squared.

toothpaste666
oops sorry L = 2(pi/2)r = pi r
so r = L/pi
so kQ/r = kQpi/L

mfb
Mentor
Right.

toothpaste666
thanks guys

i feel like i did something wrong. if my integral was k∫(Q/L)dL/r then wouldnt it be
k∫(Q/L)dL/r = kQ/r∫dL/L(because Q and r are constant) = kQ/r (ln(L)) ??

or If i put everything in terms of L instead of r from the begining:
k∫((Q/L)dL)/(L/pi) = k∫piQdL/L^2 = kpiQ∫dL/L^2 = - kpiQ/L
so it would be negative?

mfb
Mentor
If you use L like this, you have to integrate over a different variable, like L' (or x, or whatever). The L in Q/L does not change in the integral.

the L in r = L/pi changes? How would I have to set that up for it to be correct?

mfb
Mentor
No it does not change!
That's why it is better to use a different integration variable along the length of the semicircle, like x.