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(D) The length of the wire is $L = 20 \, cm = 0.2 \, m$. The wire forms a semicircle,so $\pi r = L$,which gives the radius $r = L/\pi = 0.2/\pi \, m$.

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$(25 	imes 10^3 \, N/C) \hat{i}$

Vedclass · Answer

(D) The length of the wire is $L = 20 \, cm = 0.2 \, m$. The wire forms a semicircle,so $\pi r = L$,which gives the radius $r = L/\pi = 0.2/\pi \, m$.Each half of the arc has length $L/2 = \pi r / 2$. The linear charge density is $\lambda = \pm Q / (L/2) = \pm 2Q / L$.The electric field at the center of a quarter-circular arc of charge $Q$ and radius $r$ is $E = \frac{\sqrt{2} K \lambda}{r}$ at an angle of $45^\circ$ to the symmetry axis.For the left quarter (charge $+Q$),the field $E_1$ points away from the arc at $45^\circ$ to the $x$ and $y$ axes: $E_1 = \frac{\sqrt{2} K (2Q/L)}{r} (\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = \frac{2KQ}{Lr} (\hat{i} + \hat{j})$.For the right quarter (charge $-Q$),the field $E_2$ points towards the arc at $45^\circ$ to the $x$ and $y$ axes: $E_2 = \frac{\sqrt{2} K (2Q/L)}{r} (\cos 45^\circ \hat{i} - \sin 45^\circ \hat{j}) = \frac{2KQ}{Lr} (\hat{i} - \hat{j})$.The net field $E = E_1 + E_2 = \frac{4KQ}{Lr} \hat{i} = \frac{4KQ}{L(L/\pi)} \hat{i} = \frac{4\pi KQ}{L^2} \hat{i}$.Substituting $K = 1/(4\pi\varepsilon_0)$ and $Q = 10^3 \varepsilon_0$:$E = \frac{4\pi (1/4\pi\varepsilon_0) (10^3 \varepsilon_0)}{(0.2)^2} \hat{i} = \frac{10^3}{0.04} \hat{i} = 25 	imes 10^3 \, N/C \hat{i}$.

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