Question

The rod forms a 10/22 of circle with radius R and produces an electric field of magnitude Earc at its center of curvature P.

Answer 1

Answer:

$\frac{E_q}{E_a}=\frac{\pi}{2}$

Explanation:

Assuming this problem: "Part (a) of the figure attached shows a non-conducting rod with a uniformly distributed charge +Q. The rod forms a half circle with radius R and produces an electric field of magnitude Earc at its center of curvature P. If the arc is collapsed to a point at distance R from P (Figure b), by what factor is the magnitude of the electric field at P multiplied?"

On this case the charge density is given by this formula:

$\lambda =\frac{Q}{\pi R}$ assuming a half circle

We can find the force acting on the x axis with this:

$dF_x = k \int \frac{dq}{R^2} cos \theta = \frac{K}{R^2}\int 2R d \theta cos \theta$

$dF_x= - \frac{K}{R^2} (\frac{Q}{\pi R}) R \int_{-\pi/2}^{\pi/2} cos \theta d \theta$

We can cnvert the integral using the symmetrical property:

$dF_x = \frac{KQ}{\pi R^2} 2 \int_{0}^{\pi/2} cos \theta d \theta$

And we can find the electric field like this:

$E_{a}=\frac{2KQ}{\pi R^2}$

And the electric field just by the charge is given by:

$E_q = \frac{KQ}{R^2}$

And if we find the ratio for the two electrical fields we got:

$\frac{E_q}{E_a}=\frac{\pi}{2}$