Question

Figure shows two equipotential lines in $\text{XY}$ plane for an electric field. The scales are marked. The $\text{X-}$component $E_x$ and $\text{Y-}$component $E_y$ of the electric field in the space between these equipotential lines are respectively.

• A. $+100{\text{Vm}}^{-1}$, $-200{\text{Vm}}^{-1}$
• B. $+200{\text{Vm}}^{-1}$, $+100{\text{Vm}}^{-1}$
• C. $-100{\text{Vm}}^{-1}$, $+200{\text{Vm}}^{-1}$
• D. $-200{\text{Vm}}^{-1}$, $+100{\text{Vm}}^{-1}$

Answer 1

The correct option is C $-100{\text{Vm}}^{-1}$, $+200{\text{Vm}}^{-1}$


The parallel equipotential lines represent presence of a uniform Electric field in the region.

Electric field will be $E$ perpendicular to the equipotential surfaces as shown in figure.

Slope of equipotential surfaces ,
$\tan \theta =\frac{1-0}{4-2}=\frac{1}{2}$



$\Rightarrow \sin \theta =\frac{1}{\sqrt{5}}$ and $\cos \theta =\frac{2}{\sqrt{5}}$

Here, potential difference $\Delta V=2\text{volt}$

Along the direction of Electric field.
Distance between two equipotential lines, $d=0.02\sin \theta =\frac{0.2}{\sqrt{5}}$

Using, $\Delta V=Ed$ we have,

$2=E\times \frac{0.02}{\sqrt{5}}$

$\therefore E=100\sqrt{5}\text{V/m}$

From the geometry shown in figure ,
$E_x=-E\sin \theta =-100\sqrt{5}\times \frac{1}{\sqrt{5}}$

$\therefore E_x=-100\text{V/m}$

similary ,
$E_y=E\cos \theta =100\sqrt{5}\times \frac{2}{\sqrt{5}}$

$\therefore E_y=200\text{V/m}$

Hence, option (c) is the correct answer.

Why this question? In such questions the parallel equipotential surface gives hint for uniform $\overrightarrow{E}$, for which we can apply $\Delta V=Ed$ and distance between equipotential surface (d) in direction of $\overrightarrow{E}$ can be calculated by observing geometry.