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(C) The parallel equipotential lines represent a uniform electric field in the region.The electric field $\vec{E}$ is perpendicular to the equipotenti

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$-100 \ V/m, +200 \ V/m$

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(C) The parallel equipotential lines represent a uniform electric field in the region.The electric field $\vec{E}$ is perpendicular to the equipotential surfaces and points from higher potential to lower potential.The slope of the equipotential lines is $m = \frac{\Delta y}{\Delta x} = \frac{2-1}{6-4} = \frac{1}{2}$.Let $	heta$ be the angle the equipotential line makes with the $X$-axis. Then $	an 	heta = \frac{1}{2}$.This implies $\sin 	heta = \frac{1}{\sqrt{5}}$ and $\cos 	heta = \frac{2}{\sqrt{5}}$.The perpendicular distance $d$ between the two equipotential lines (potential difference $\Delta V = 4 \ V - 2 \ V = 2 \ V$) is calculated using the geometry. The horizontal distance between the lines at a constant $Y$ is $\Delta x = 6 - 4 = 2 \ cm = 0.02 \ m$.Thus,$d = \Delta x \sin 	heta = 0.02 	imes \frac{1}{\sqrt{5}} = \frac{0.02}{\sqrt{5}} \ m$.Using $\Delta V = E d$,we get $2 = E 	imes \frac{0.02}{\sqrt{5}}$,which gives $E = \frac{2 \sqrt{5}}{0.02} = 100 \sqrt{5} \ V/m$.The electric field vector points from $4 \ V$ to $2 \ V$ (towards the origin). The angle $\alpha$ that the electric field vector makes with the negative $X$-axis is $	heta$. Thus,$E_x = -E \sin 	heta = -(100 \sqrt{5}) 	imes \frac{1}{\sqrt{5}} = -100 \ V/m$.And $E_y = E \cos 	heta = (100 \sqrt{5}) 	imes \frac{2}{\sqrt{5}} = 200 \ V/m$.Therefore,$E_x = -100 \ V/m$ and $E_y = 200 \ V/m$. The correct option is $(C)$.

The figure shows two equipotential lines in the $XY$ plane for an electric field. The scale is shown. The $X$-component $E_x$ and $Y$-component $E_y$ of the electric field in the space between the equipotential lines are respectively:

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