Question

A uniform electric field pointing in positive $x$-direction exists in a region. Let $A$ be the origin, $B$ be the point on the $x$-axis at $x=+1\text{cm}$ and $C$ be the point on the $y$-axis at $y=+1\text{cm}$. Then, the potentials at the points $A,B$ and $C$ satisfy

• A. $V_A<V_B$
• B. $V_A>V_B$
• C. $V_A<V_C$
• D. $V_A>V_C$

Answer 1

The correct option is B $V_A>V_B$


Let the potential of point $A,B$ and $C$ be $V_A,V_B$ and $V_C$
As shown in figure $\overrightarrow{E}$ is along positive $x$- axis.

Equipotential surfaces are perpendicular to electric field, and are represented by dotted lines.

From the definition of an equipotential surface we can say that , $V_A=V_C$ ( as point $A$ and $C$ are on same equipotential surface)

As we move along the direction of electric field, the electric potential decreases.
So,
$V_A>V_B$

Hence, option (b) is the correct answer.