Question

Some equipotential surfaces are shown in figure. The magnitude and direction of the electric field is ($\theta = 30^\circ$)

• A. 100 V/m making angle $120^\circ$ with positive x axis
• B. 200 V/m making an angle $60^\circ$ with positive x axis
• C. 200 V/m making an angle $120^\circ$ with positive x axis
• D. 100 V/m making an angle $60^\circ$ with positive x axis

Answer 1

Answer: (C)

200 V/m making an angle $120^\circ$ with positive x axis

Answer 2

The electric field is perpendicular to the equipotential surfaces and points in the direction of decreasing potential. From the figure, the equipotential surfaces are parallel lines. The potential increases from 20 V to 50 V as we move from left to right across the lines. The angle between the equipotential lines and the positive x-axis is given as $\theta = 30^\circ$. From the figure, the equipotential lines are oriented such that they make an angle of $180^\circ - 30^\circ = 150^\circ$ with the positive x-axis.

The direction of the electric field is perpendicular to the equipotential lines. So, the angle of the electric field with the positive x-axis is $150^\circ \pm 90^\circ$. This gives angles $150^\circ - 90^\circ = 60^\circ$ or $150^\circ + 90^\circ = 240^\circ$.

The potential increases as we move from left to right. The electric field points in the direction of decreasing potential, which is from right to left, perpendicular to the equipotential lines.