Question

The equation of an equipotential region (line) is $y=\frac{3x}{4}$, then electric field vector at the origin may be

• A. $(-6\hat{i}+8\hat{j})$
• B. $(6\hat{i}+8\hat{j})$
• C. $(4\hat{i}+3\hat{j})$
• D. $(3\hat{i}+3\hat{j})$

Answer 1

The correct option is A $(-6\hat{i}+8\hat{j})$


Equation of equipotential surface is $y=\frac{3x}{4}$

Now, electric field along the euipotential surface should be zero, therefore angle made by the equipotential surface with the x-axis is
$\tan \theta =\frac{3}{4}$

Now, since net electric field should be perpendicular to the equipotential surface, therefore for any electric field which makes an angle

$\tan (90-\theta )=\frac{E_y}{E_x}=\frac{-4}{3}$
$\frac{-4}{3}=\frac{E_y}{E_x}$
$E_y=\frac{-4E_x}{3}$

Hence, electric field vector at origin can be $(-6\hat{i}+8\hat{j})$.