Answer to Question #104795 in Electricity and Magnetism for havefun7741

Question

[img]https://upload.cc/i1/2020/03/06/7mitHn.jpg[/img]



R is 4 in the question

Answer 1

As per the given question,

Charge on the thicker part of the ring$= +20\mu C$

Radius is =R+3

Charge on the thinner part of the ring $=-40\mu C$

Radius = R+2

$\epsilon_o=8.85\times 10^{-12} C^2/N-m^2$

Let arc dx is making $d\theta$ at the origin,

Let the electric field due to the thicker charged ring is $E_1$ and due to the thinner part is $E_2$

$dq= \dfrac{20dx}{\dfrac{\pi}{2}(R+3)}=\dfrac{40 dx}{\pi(R+3)}$

$dE_1=\dfrac{dq}{4\pi^2 \epsilon_o (R+3)^2}=\dfrac{40dx}{4\pi^2\epsilon_o (R+3)^3}$

$\Rightarrow dE_1=\dfrac{40(R+3)d\theta}{4\pi^2 \epsilon_o(R+3)^3}$

Now taking integration of both side $\int^{E}_{0} dE_1=\int^{\pi/2}_{\pi}\dfrac{40(R+3)d\theta}{4\pi^2 \epsilon_o(R+3)^3}$

$\int^{E}_{0} dE_1=\int^{\pi/2}_{\pi}\dfrac{40d\theta}{4\pi^2\epsilon_o (R+3)^2}=\dfrac{5}{\pi\epsilon_o(R+3)^2}$ $(\hat{i}-\hat{j})$

Similarly for the thinner arc

$\int^{E_2}_{0} dE_2=\int^{\pi/2}_{\pi}\dfrac{40d\theta}{4\pi^2 (R+2)^2}=\dfrac{10}{\pi\epsilon_o(R+2)^2}$

$E_2=\dfrac{10}{\pi(\epsilon_o(R+2)^2)}(\hat{i})$

Hence the resultant electric field,

$E_{net}=(\dfrac{5}{\pi\epsilon_o(R+3)^2}+\dfrac{10}{\pi(\epsilon_o(R+2)^2)})(\hat{i}) -\dfrac{5}{\pi\epsilon_o(R+3)^2}\hat{j}$