Physics · Electrostatic Potential And Capacitance · NEET
Because the electrostatic (Coulomb) force is a conservative force. For any conservative force, the work done depends only on the initial and final positions, not on the route between them. NCERT states this directly for gravity in Class 11 and extends the same idea to the electric force in Class 12: the work equals the change in potential energy, W = U_B − U_A = q(V_B − V_A). Since V_A and V_B are fixed once you fix the two points, every path gives the same work.
They are the same physical fact stated two ways. 'Conservative force' is the cause; 'work depends only on end points' is the result. A third equivalent way: the work done in taking a charge around any closed loop is zero. If any one of these is true, all three are true for the electrostatic field.
Use W = q·ΔV = q(V_B − V_A), the potential difference between the end points. Writing W = qV secretly means V is measured with respect to infinity (where V = 0). Since infinity is fixed, that is still a difference. For NEET, always subtract the two potentials of the actual start and end points.
No. It is zero only if the start and end points are on the same equipotential surface (V_B = V_A), or if you return to the start (closed loop). If V_B ≠ V_A, the work is a fixed non-zero number, the same for every path, but not zero.
On an equipotential surface every point has the same potential, so V_B − V_A = 0 and W = q·0 = 0. Also, the field is always perpendicular to the equipotential surface, so the force has no component along your direction of motion, meaning zero work.
Four diagrams show regions of equipotentials (10 V, 20 V, 30 V, 40 V) arranged differently. A positive charge is moved from A to B in each diagram, where A and B lie on the 10 V and 40 V equipotentials. Which statement is correct?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. The electric field of any set of static (fixed) charges is conservative, so work done on a test charge is always path-independent. This is a core property used throughout the potential chapter.
Friction is a non-conservative force: the work it does depends on the path length, so a longer path loses more energy. The electrostatic force is conservative, so path length does not matter, only the end points do.
W_external = q(V_B − V_A) to move the charge slowly (no change in kinetic energy). The work done by the field itself is the negative of this: W_field = q(V_A − V_B).
A closed loop has the same start and end point, so V_B = V_A and ΔV = 0, giving W = 0. This zero-loop property is the defining test of a conservative field.
It changes the sign and size of the work through the q term, but it does not change the fact that the work is path-independent. Only the two end-point potentials and the charge value matter.
