The diagram shows symmetrically placed rectan | vedclass

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(C) $1$. The positively charged rectangular insulator on the right produces an electric field $\vec{E}_+$ at the origin pointing away from it,directed

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aligned with the positive $y$-axis.

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(C) $1$. The positively charged rectangular insulator on the right produces an electric field $\vec{E}_+$ at the origin pointing away from it,directed into the second quadrant.$2$. The negatively charged rectangular insulator on the left produces an electric field $\vec{E}_-$ at the origin pointing towards it,also directed into the second quadrant.$3$. Since the insulators are symmetrically placed and have equal charge magnitudes,the magnitudes of these electric fields are equal,i.e.,$|\vec{E}_+| = |\vec{E}_-| = E$.$4$. Both fields have components along the negative $x$-axis and the positive $y$-axis. Specifically,if $	heta$ is the angle each field makes with the $x$-axis,the $x$-components are $-E \cos 	heta$ and the $y$-components are $E \sin 	heta$.$5$. The net electric field $\vec{E}_{net} = \vec{E}_+ + \vec{E}_-$ will have a horizontal component of $-2E \cos 	heta$ (along the negative $x$-axis) and a vertical component of $2E \sin 	heta$ (along the positive $y$-axis).$6$. However,looking at the geometry,the electric field from the positive plate points away from the plate (towards the second quadrant),and the field from the negative plate points towards the plate (also towards the second quadrant). The vector sum of these two fields results in a net field directed along the positive $y$-axis because the $x$-components cancel out due to symmetry if the plates are oriented such that their fields are symmetric about the $y$-axis.

The diagram shows symmetrically placed rectangular insulators with uniformly charged distributions of equal magnitude. At the origin,the net electric field $\vec{E}_{net}$ is:

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