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(A) Consider a small element of the arc at an angle $	heta$ with the vertical,having an angular width $d	heta$. The charge on this element is $dq =

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) Consider a small element of the arc at an angle $	heta$ with the vertical,having an angular width $d	heta$. The charge on this element is $dq = \lambda (R d	heta) = (\lambda_0 \sin 	heta) R d	heta$.The electric field due to this element at the center $P$ is $dE = \frac{k dq}{R^2} = \frac{1}{4\pi \epsilon_0} \frac{\lambda_0 \sin 	heta R d	heta}{R^2} = \frac{\lambda_0 \sin 	heta d	heta}{4\pi \epsilon_0 R}$.By symmetry,the horizontal components of the electric field cancel out. The net electric field $E_1$ is directed along the vertical axis and is the integral of the vertical components $dE \cos 	heta$ from $	heta = 0$ to $	heta = \pi$ (or $	heta = -\pi/2$ to $\pi/2$ depending on the reference axis; based on the figure,we integrate from $0$ to $\pi$):$E_1 = \int_0^{\pi} dE \cos 	heta = \int_0^{\pi} \frac{\lambda_0 \sin 	heta \cos 	heta d	heta}{4\pi \epsilon_0 R} = \frac{\lambda_0}{4\pi \epsilon_0 R} \int_0^{\pi} \sin 	heta \cos 	heta d	heta$.Using $\sin 	heta \cos 	heta = \frac{1}{2} \sin(2	heta)$:$E_1 = \frac{\lambda_0}{8\pi \epsilon_0 R} \int_0^{\pi} \sin(2	heta) d	heta = \frac{\lambda_0}{8\pi \epsilon_0 R} \left[ -\frac{\cos(2	heta)}{2} ight]_0^{\pi} = \frac{\lambda_0}{8\pi \epsilon_0 R} \left( -\frac{1}{2} + \frac{1}{2} ight) = 0$.Wait,re-evaluating based on the provided solution image logic where $E_x = \int dE \sin 	heta$ and the integration limits are $-\pi/2$ to $\pi/2$:$E_1 = \int_{-\pi/2}^{\pi/2} \frac{\lambda_0 \sin 	heta d	heta}{4\pi \epsilon_0 R} \sin 	heta = \frac{\lambda_0}{4\pi \epsilon_0 R} \int_{-\pi/2}^{\pi/2} \sin^2 	heta d	heta = \frac{\lambda_0}{4\pi \epsilon_0 R} \left[ \frac{	heta}{2} - \frac{\sin(2	heta)}{4} ight]_{-\pi/2}^{\pi/2} = \frac{\lambda_0}{4\pi \epsilon_0 R} \left( \frac{\pi}{4} - (-\frac{\pi}{4}) ight) = \frac{\lambda_0}{4\pi \epsilon_0 R} \cdot \frac{\pi}{2} = \frac{\lambda_0}{8 \epsilon_0 R}$.Thus,$\frac{\lambda_0}{\epsilon_0 E_1 R} = \frac{\lambda_0}{\epsilon_0 (\lambda_0 / 8 \epsilon_0 R) R} = 8$.

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