Question

Electric Field:

You are helping to design a new electron microscope to investigate the

structure of the HIV virus. A new device to posit

ion the electron beam consists of a charged

circle of conductor. This circle is divided into two half circles separated by a thin insulator

so that half of the circle can be charged positively and half can be charged negatively. The

electron beam will go

through the center of the circle. To complete the design your job is to

calculate the electric field in the center of the circle as a function of the amount of positive

charge on the half circle, the amount of negative charge on the half circle, and the

radius of

the circle.

Answer 1

Let the electric field at the center of the charged ring be $E_1$ due to the positively charge part and electric field at the center due to the negatively charged ring be $E_2$

Let the charge density of the ring be $\lambda$

So net positive charge on ring $=\pi a\lambda$

Electric field along y axis $dE_1=\frac{\lambda ad\theta \cos\theta }{4\pi \epsilon a^2}$

Now, taking integration of both side,

$\int dE_1=\int_{-\pi/2}^{\pi/2}\frac{\lambda ad\theta \cos\theta }{4\pi \epsilon a^2}\hat{j}$

$=\frac{\lambda }{4\pi \epsilon a}[\sin\theta]_{\pi/2}^{-\pi/2}\hat{j}$

$=\frac{\lambda }{4\pi \epsilon a}[\sin(\pi/2)-\sin(-\pi/2)]\hat{j}$

$=\frac{2\lambda }{4\pi \epsilon a}\hat{j}$

$=\frac{\lambda }{2\pi \epsilon a}\hat{j}$

Total negative charge on the ring $=-\pi a\lambda$

So, electric field due to the negative charge $E_2=\frac{-2\lambda }{4\pi \epsilon a}\hat{-j}$

Hence, the net electric field $E=E_1+E_2$

$=0$

Hence, net electric field at the center will be zero.