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(A) The electric field $E$ is perpendicular to the equipotential surfaces and points from higher potential to lower potential.From the figure,the equi

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$10 \sqrt{2} \, V/m$ at $45^{\circ}$ with the $x$-axis

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(A) The electric field $E$ is perpendicular to the equipotential surfaces and points from higher potential to lower potential.From the figure,the equipotential lines make an angle of $135^{\circ}$ with the positive $x$-axis.The normal to these lines makes an angle of $135^{\circ} - 90^{\circ} = 45^{\circ}$ with the positive $x$-axis.The potential difference between two consecutive surfaces is $\Delta V = 10 \, V$.The perpendicular distance $d$ between two consecutive surfaces can be calculated from the geometry: $d = 1 \cdot \sin(45^{\circ}) = \frac{1}{\sqrt{2}} \, m$.The magnitude of the electric field is $E = \frac{|\Delta V|}{d} = \frac{10}{1/\sqrt{2}} = 10 \sqrt{2} \, V/m$.Since the field points from higher potential to lower potential,it is directed at $45^{\circ}$ with the $x$-axis.

The figure shows a set of equipotential surfaces. The magnitude and direction of the electric field that exists in the region is .........

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