Question

Figure shows, two equipotential surfaces $V_1$ and $V_2$. The component of electric field (in SI units) along $x$ and $y$ direction will be -

• A. $100\hat{i}\text{and}100\hat{j}$
• B. $-100\hat{i}\text{and}100\hat{j}$
• C. $100\hat{i}\text{and}-100\hat{j}$
• D. $-100\hat{i}\text{and}-100\hat{j}$

Answer 1

The correct option is B $-100\hat{i}\text{and}100\hat{j}$

Electric field lines are perpendicular to the equipotential surface and are directed from high potential to low potential.

$\therefore$ Electric field will have one component in negative $x$ direction and one component in $+y$ direction.
$E_x=-\frac{\Delta V}{\Delta x}=\frac{-2}{0.02}=-100{\text{Vm}}^{-1}$
$E_y=-\frac{\Delta V}{\Delta y}=\frac{2}{0.02}=100{\text{Vm}}^{-1}$