$A$ circle of radius $R$ is drawn in a uniform electric field as shown in the figure. If the potentials at points $A, B, C,$ and $D$ on the circumference of the circle are $V_A, V_B, V_C,$ and $V_D$ respectively,then:

Show that the direction of the electric field at a given point is normal to the equipotential surface passing through that point.

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

Two charges $2 \; \mu C$ and $-2 \; \mu C$ are placed at points $A$ and $B$ which are $6 \; cm$ apart.
$(a)$ Identify an equipotential surface of the system.
$(b)$ What is the direction of the electric field at every point on this surface?

$A$ uniform electric field exists in the plane of the paper as shown. Here $A, B, C$, and $D$ are points on the circle. $V_{A}, V_{B}, V_{C}$, and $V_{D}$ are the potentials at those points respectively. Then:

This question has Statement-$1$ and Statement-$2$. Of the four choices given after the Statements,choose the one that best describes the two Statements.
Statement-$1$: No work is required to be done to move a test charge between any two points on an equipotential surface.
Statement-$2$: Electric lines of force at the equipotential surfaces are mutually perpendicular to each other.