Answer to Question #237830 in Physics for Ismah

Question

A ring of radius 2cm carries a uniformly distributed positive total charge of 2microcoulombs . calculate the electric field at a point P, lying 3cm from it center along the central axis perpendicular to the plane of the ring.

Answer 1

Let’s consider the small charge element $dq$ on the ring. The total field $dE$ at point $P$ is the superposition of all fields due to charge elements around the ring. The electric field $dE$ has two components: $dE_x=dEcos\theta$ along the $x$-axis and $dE_y=dEsin\theta$ along the y-axis. During the integration around the ring the non-$x$ components will be cancelled by the symmetry. Thus, we get only the $x$-component:

$$dE_x=dEcos\theta=\dfrac{kdq}{r^2}cos\theta.$$

We can find $cos\theta$ from the geometry of the task:

$$cos\theta=\dfrac{x}{r}.$$

Therefore, we can write:

$$dE_x=\dfrac{kdq}{r^2}\dfrac{x}{r}.$$

We can find find $r$ from the Pythagorean theorem:

$$r=\sqrt{x^2+a^2},$$

here, $a$ is the radius of the ring.

Finally, we get:

$$dE_x=\dfrac{kdq}{(x^2+a^2)}\dfrac{x}{\sqrt{x^2+a^2}}=\dfrac{kx}{(x^2+a^2)^{3/2}}dq.$$

After integrating, we get:

$$E_x= \int\dfrac{kx}{(x^2+a^2)^{3/2}}dq,$$ $$E_x= \dfrac{kx}{(x^2+a^2)^{3/2}} \int dq=\dfrac{kx}{(x^2+a^2)^{3/2}}Q.$$

Finally, we can calculate the electric field at a point $P$ lying 3 cm from it center along the central axis perpendicular to the plane of the ring:

$$E_x(x=0.03\ m)=\dfrac{8.99\cdot10^9\ \dfrac{N\cdot m^2}{C^2}\cdot 0.03\ m}{((0.03\ m)^2+(0.02\ m)^2)^{3/2}}\cdot2\cdot10^{-6}\ C,$$ $$E_x(x=0.03\ m)=1.15\cdot10^7\ \dfrac{N}{C}.$$