Physics · Electrostatic Potential And Capacitance · NEET
Suppose the field made an angle other than 90 degrees with the surface. Then it would have a component lying along the surface. That component would push a test charge sideways and do work: W = component of force times distance. But moving along an equipotential means V is constant, so the potential difference is zero and W = q times 0 = 0. A non-zero side component cannot give zero work. The only way to have zero work is for the field to have no component along the surface, which means it points exactly perpendicular (90 degrees).
Work to move a charge between two points is W = q(V_final - V_initial). On one equipotential surface every point has the same V, so V_final - V_initial = 0. Therefore W = 0 for any path staying on the surface. Since work equals force times displacement in the direction of motion, zero work means the force (and hence the field) has zero component along the surface.
The relation E = -dV/dr says the field points where potential drops fastest (the steepest slope). Along an equipotential, V does not change at all, so dV = 0 in that direction, giving zero field component there. The full field therefore points in the perpendicular direction, where V changes fastest. This is exactly the topic of the next page, relation between electric field and potential.
It holds for every charge configuration. NCERT proves it generally: for a point charge the equipotentials are concentric spheres and the radial field is normal to them, but the 'no work along the surface' argument does not depend on the shape. So for a dipole, a plate, a conductor, or any arrangement, the field at any point is perpendicular to the equipotential through that point.
Because the field magnitude equals the potential change per unit perpendicular distance: E = dV / dl (taking magnitude). For a fixed step in V between two surfaces, a larger E means a smaller distance dl. So closely packed equipotentials indicate a strong field, and widely spaced ones indicate a weak field.
The angle between the electric lines of force and the equipotential surface is:
Diagrams (a), (b), (c), (d) show regions of equipotentials (values 10 V, 20 V, 30 V, 40 V arranged differently). A positive charge q is moved from A to B in each diagram. Which statement is correct?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Always 90 degrees. The field is normal (perpendicular) to the equipotential surface at every point, for any charge arrangement.
No. Since V is the same at all points on the surface, the potential difference is zero, so W = q times 0 = 0, no matter which path you take on the surface.
It would have a component along the surface that does work on a moving charge. That work would change the potential, contradicting the definition of an equipotential surface. So the field must be perpendicular.
Along the surface V does not change, so dV = 0 and the field component there is zero. The field points only in the direction of steepest V change, which is perpendicular to the surface.
It is a direct one-mark question (asked in NEET 2022) and it underpins problems on work done between equipotentials (NEET 2017). Knowing the 90-degree rule and 'zero work along the surface' solves both instantly.
