Question

Four separate arcs of a circle are joined to form a new circle, as shown. The black arc has a charge +Q uniformly distributed over its length. The three remaining shaded arcs each have -Q similarly distributed.

If the +Q charge produces an electric field of 1.5 N/C at the center of the circle, find the net field at the center due to all four arcs.

magnitude
______N/C
direction?

lower right upper left     upper right in right down lower left up left out

Answer 1

E due to part in 2nd and 4th quadrant will cancel each other as both will apply same E at center in opposite direction

but in 1st quadrant and 3rd quadrant ,both part have - and + charge respectively so both will apply same E in same direction so net E at center is 1.5+1.5 = 3 N/C

direction= upper right

Answer 2

Considereing the black quater, if a small element dL is considered on this quater. Electric field dE = k x dQ/r² where r is the radius. Total charge = 2Q/(pi r). dQ = 2QdL/(pi r).

$E = \int_{\pi}^{3\pi/2} k 2Q/\pi r^{3} (cos\theta \hat{i} + sin\theta \hat{j}) dL\\ = \int_{\pi}^{3\pi/2} k 2Q/\pi r^{3} (cos\theta \hat{i} + sin\theta \hat{j}) rd\theta\\ = [\frac{2kQ}{\pi r^{2}} (-sin \theta \hat{i} + cos \theta \hat{j})]_{\pi\rightarrow \3pi/2}\\ E = \frac{2kQ}{\pi r^{2}} (\frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}})\\ E = \frac{\sqrt{2}kQ}{\pi r^{2}} (\hat{i}+ \hat{j})$

Field due the the 2nd quadrant and 4th quadrant will be cancelled as they are equal and opposite.

FIeld due to the part in the first quadrant and 3rd quadrant are equal in magnitude but in the same direction that is (i + j). Therefore, the net field is

E(total) = 2 x 1.5 = 3 N/C (i + j)( upper right)