Question

1. (5 points) A semi-annulus with inner radius rı and outer radius r2 is placed on the ry plane at z 0, with centre of the radii at the origin, sllustrated. The half-annulus has a uniform surface charge density ơ r 2 a) Find the potential V at the origin. b) Find E at the origin. (Can you use the result of a) to get E?)

Answer 1

Solution

a)

Potential at the center for half annulus ring can be given by Potential due to half disk of radius r₂- potential due to half disk of radius r₁

$V_{total}=V_{r_{2}}-V_{r_{1}}$

we know for a semi circular disk of radius r and surface charge density $\sigma$ is given by

$V=k\pi \sigma r$

$\therefore V_{total}=k\pi \sigma \left (r_{2}-r_{1} \right )$

b)

To find electric field

We assume a thin semi circular ring of radius a and thickness da from the center and charge dQ

we know,

Electric field due to semi circular ring of radius 'a' is given by,

$\frac{Q}{2\pi ^{2}a^{2}\epsilon _{o}}$

This can also be derived by taking a semi circular ring and find the electric field by integrating assumed point charges on the rod. proof given below

Now,

For the ring assumed

$\mathrm{d}E=\frac{\mathrm{d}Q}{2\pi ^{2}a^{2}\epsilon _{o}}$

and

$\mathrm{d}Q=\sigma \pi a \mathrm{d}a$

So, to get the electric field of the semi annular disk.

$\mathrm{d}E=\frac{\sigma \pi a \mathrm{d}a}{2\pi ^{2}a^{2}\epsilon _{o}}$

$\Rightarrow \mathrm{d}E=\frac{\sigma \mathrm{d}a}{2\pi a \epsilon _{o}}$

$\Rightarrow \oint \mathrm{d}E=\int_{r_{1}}^{r_{2}}\frac{\sigma \mathrm{d}a}{2\pi a \epsilon _{o}}$

$\Rightarrow E=\left |\frac{\sigma \ln a}{2\pi \epsilon _{o}} \right |_{r_{1}}^{r_{2}}$

$\Rightarrow E=\frac{\sigma \ln r_{2}-\ln r_{1} }{2\pi \epsilon _{o}}$

$\Rightarrow E=\frac{\sigma }{2\pi \epsilon _{o}}\ln \frac{r_{2}}{r_{1}}$

No, we cannot directly use the result of a) to get b) because E=dV/dr is applicable for point charges only. For these kind of bodies we need to find potential of its elements and then integrate it to get the answer.

If any doubt feel free to comment.