What is the electric potential at the center of a ring of charge?

[Necraz]

[SOLVED] Voltage from Ring of Charge

Homework Statement

A plastic rod has been bent into a circle of radius R = 6.40 cm. It has a charge Q₁ = +2.40 pC uniformly distributed along one-quarter of its circumference and a charge Q₂ = -6Q₁ uniformly distributed along the rest of the circumference (Figure 24-39). Take V = 0 at infinity. What is the electric potential at the center C of the circle?

Homework Equations

dV = \frac{dq}{4\pi\epsilon_0r}
\lambda = \frac{q}{L}

The Attempt at a Solution

dq = \lambda r d \theta
\lambda = \frac{q}{2 \pi r}
V = \int_{0}^{\phi}{\frac{\lambda d \theta}{4\pi\epsilon_0}} = \frac{\lambda \phi}{4 \pi \epsilon_0} = \frac{q \phi}{8 \pi^2 \epsilon_0 r}
V_1 = \frac{(2.4*10^-12)(\pi / 2)}{8 \pi^2 \epsilon_0 (0.064)} = 0.0843V
V_2 = \frac{-6(2.4*10^-12)(3\pi / 2)}{8 \pi^2 \epsilon_0 (0.064)} = -1.517V
V_{net} = V_1 + V_2 = -1.433V (Incorrect)

 

[Oerg]

You should start from integrating the Differential electric field strength that a point charge would exert at that point

 

[Necraz]

You should start from integrating the Differential electric field strength that a point charge would exert at that point

Field:
\lambda = \frac{q}{L} = \frac{q}{2 \pi r} = \frac{dq}{ds}
ds = r d\theta
dq = \lambda r d\theta
dE = \frac{cos\theta dq}{4 \pi \epsilon_0 r^2}
E = \int_{\frac{-\phi}{2}}^{\frac{\phi}{2}}{\frac{\lambda cos\theta d\theta}{4 \pi \epsilon_0 r}} = \frac{\lambda(sin\frac{\phi}{2} - sin\frac{-\phi}{2})}{4 \pi \epsilon_0 r} = \frac{qsin\frac{\phi}{2}}{4 \pi^2 \epsilon_0 r^2}

Potential:
E = \frac{dV}{dr}
V = \int{\frac{qsin\frac{\phi}{2}dr}{4 \pi^2 \epsilon_0 r^2}} = \frac{q sin\frac{\phi}{2}}{4 \pi^2 \epsilon_0} \int{\frac{dr}{r^2}} = \frac{-q sin\frac{\phi}{2}}{4 \pi^2 \epsilon_0 r}

Values:
V_1 = \frac{-2.4*10^{-12} sin\frac{\pi/2}{2}}{4 \pi^2 \epsilon_0 (0.064)} = -0.076V
V_2 = \frac{6 * 2.4*10^{-12} sin\frac{3\pi/2}{2}}{4 \pi^2 \epsilon_0 (0.064)} = 0.455V
V_{net} = V_1 + V_2 = 0.379 V (Incorrect)

 

[kamerling]

your expression for \lambda is wrong. the charge isn't distributed along the whole circle, but along a secion of 1/4 or 3/4 of the circle. The rest is OK.

The whole problem gets much simpler, if you see that V doesn't depend on where the charge is, but only on the distance. Since all the charges are at the same distance from the center, you can move them all to a single point at distance r from the cednter

 

[Necraz]

your expression for \lambda is wrong. the charge isn't distributed along the whole circle, but along a secion of 1/4 or 3/4 of the circle. The rest is OK.

Hadn't even considered that the density equation could be wrong - I figured the mistake was somewhere in the following page of calculus. Thanks for pointing that out, that must be why the answers are incorrect.

The whole problem gets much simpler, if you see that V doesn't depend on where the charge is, but only on the distance. Since all the charges are at the same distance from the center, you can move them all to a single point at distance r from the cednter

Realized this and got the answer correct just prior to returning here to mark this thread solved. Thanks for your help!