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(B) The electric field due to a charged arc at its center is directed along the angle bisector. For arc $AB$ (a quarter circle),the electric field $E$

Vedclass · Accepted Answer

$\sqrt{2} E$

Vedclass · Answer

(B) The electric field due to a charged arc at its center is directed along the angle bisector. For arc $AB$ (a quarter circle),the electric field $E$ at $O$ is directed at $45^\circ$ from the radii $OA$ and $OB$. This field $E$ can be resolved into two components: $E_x = E \cos(45^\circ) = E / \sqrt{2}$ along the negative $x$-axis (towards $D$) and $E_y = E \sin(45^\circ) = E / \sqrt{2}$ along the negative $y$-axis (towards $C$). Arc $ABC$ consists of two identical quarter-circle arcs: $AB$ and $BC$. For arc $AB$,the field is $\vec{E}_{AB} = (-E/\sqrt{2}) \hat{i} + (-E/\sqrt{2}) \hat{j}$. For arc $BC$,the field is $\vec{E}_{BC} = (E/\sqrt{2}) \hat{i} + (-E/\sqrt{2}) \hat{j}$. Adding these vectorially,the $x$-components cancel out,and the $y$-components add up: $\vec{E}_{ABC} = \vec{E}_{AB} + \vec{E}_{BC} = 0 \hat{i} + (-2E/\sqrt{2}) \hat{j} = -\sqrt{2} E \hat{j}$. The magnitude is $\sqrt{2} E$.

$A$ metallic ring is uniformly charged as shown in the figure. $AC$ and $BD$ are two mutually perpendicular diameters. The magnitude of the electric field at $O$ due to arc $AB$ is $E$. What would be the magnitude of the electric field at $O$ due to arc $ABC$?

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