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(A) For a semi-circular ring of radius $R$ and total charge $Q$,the linear charge density is $\lambda = \frac{Q}{\pi R}$.The electric field $E$ at the

Vedclass · Accepted Answer

$2.14$

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(A) For a semi-circular ring of radius $R$ and total charge $Q$,the linear charge density is $\lambda = \frac{Q}{\pi R}$.The electric field $E$ at the center is given by the formula $E = \frac{2k\lambda}{R}$,where $k = \frac{1}{4\pi\epsilon_0}$.Substituting $\lambda$,we get $E = \frac{2(1/4\pi\epsilon_0)(Q/\pi R)}{R} = \frac{Q}{2\pi^2 \epsilon_0 R^2}$.Given $E = 100 	ext{ V/m}$,$R = 0.35 	ext{ m}$,$\epsilon_0 = 8.85 	imes 10^{-12} 	ext{ C}^2/	ext{Nm}^2$,and $\pi = 3.14$.Rearranging for $Q$: $Q = E 	imes 2\pi^2 \epsilon_0 R^2$.$Q = 100 	imes 2 	imes (3.14)^2 	imes 8.85 	imes 10^{-12} 	imes (0.35)^2$.$Q = 200 	imes 9.8596 	imes 8.85 	imes 10^{-12} 	imes 0.1225$.$Q \approx 2.14 	imes 10^{-9} 	ext{ C} = 2.14 	ext{ nC}$.

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