Question

$S_1$ and $S_2$ are two equipotential surfaces on which the potentials are not equal. Then-

• A. $S_1$ and $S_2$ both cannot intersect.
• B. $S_1$ and $S_2$ both cannot be plane surfaces.
• C. In the region between $S_1$ and $S_2$, the field is maximum where they are closed to each other.
• D. A line of force from $S_1$ and $S_2$ must be perpendicular to both.

Answer 1

Answer: (A), (C), (D)

$S_1$ and $S_2$ both cannot intersect.
In the region between $S_1$ and $S_2$, the field is maximum where they are closed to each other.
A line of force from $S_1$ and $S_2$ must be perpendicular to both.

As we know that two equipotential surfaces can not intersect each other and lines of force are always perpendicular to the equipotential surface.
When two equipotential lines close together, the slope will be steep. So, close equipotential surfaces indicate a maximum electric field.$(E=-\frac{dV}{dx})$
Ans: $\left(A\right),\left(C\right),\left(D\right)$