A semicircular ring of radius 0.5 m is unifo | vedclass

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(D) The electric field $E$ at the center of a uniformly charged semicircular ring is given by the formula $E = \frac{2k\lambda}{r} \sin(\frac{\alpha}{

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(D) The electric field $E$ at the center of a uniformly charged semicircular ring is given by the formula $E = \frac{2k\lambda}{r} \sin(\frac{\alpha}{2})$,where $\alpha$ is the angle subtended at the center. For a semicircle,$\alpha = 180^\circ$,so $\sin(90^\circ) = 1$. The linear charge density $\lambda = \frac{Q}{L} = \frac{Q}{\pi r}$. Substituting these into the formula: $E = \frac{2k(Q/\pi r)}{r} = \frac{2kQ}{\pi r^2}$. Given: $k = 9 	imes 10^9 \ N \ m^2/C^2$,$Q = 1.4 	imes 10^{-9} \ C$,$r = 0.5 \ m$. $E = \frac{2 	imes (9 	imes 10^9) 	imes (1.4 	imes 10^{-9})}{\pi 	imes (0.5)^2} = \frac{25.2}{\pi 	imes 0.25} = \frac{100.8}{\pi} \approx 32.09 \ V/m$. Rounding to the nearest integer,the value is $32 \ V/m$.

$A$ semicircular ring of radius $0.5 \ m$ is uniformly charged with a total charge of $1.4 \times 10^{-9} \ C$. The electric field intensity at the center of the ring is ........ $V/m$.

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