Physics · Electrostatic Potential And Capacitance · NEET
Work done to move a charge q between two points is W = q(V_B - V_A), where V_A and V_B are the potentials at the start and end. It depends only on the two endpoints, not on the path taken. So if you move a charge along the same equipotential surface, V does not change, V_B - V_A = 0, and the work done is zero. Memory hook: "Same voltage, zero work; only the potential difference between endpoints matters."
Left: moving charge q from the 10 V line (A) to the 30 V line (B) needs W = q(30 - 10). The straight and curved paths give equal work. Right: moving along one equipotential surface gives V_B - V_A = 0, so W = 0.
Your doubts, answered
Is the work done really zero when I move a charge along an equipotential surface?
Yes. An equipotential surface is a surface where every point has the same potential V. If both the start and end points lie on it, then V_B - V_A = 0, so W = q(V_B - V_A) = 0. This is true no matter how long or curved the path is, as long as you begin and end on the same equipotential. This is why the electric field has no component along an equipotential surface.
Does the path I take change the work done?
No. Electrostatic force is a conservative force, so the work done depends only on the initial and final potentials, not on the route. Two students moving the same charge between the same two equipotentials do equal work, even if one takes a straight line and the other a zig-zag path. Only V_A and V_B decide the answer.
Should I use W = qV or W = q(V_B - V_A)?
Use W = q(V_B - V_A), the potential difference between the two endpoints. W = qV only works when you bring the charge from infinity (where V = 0), because then V_A = 0 and V_B - V_A = V. Mixing these up is a common NEET error. When endpoints are given directly, always take the difference.
What is the difference between work done BY the field and work done BY me (external agent)?
When you move the charge slowly (no change in kinetic energy), the external work you do is W_ext = q(V_B - V_A). The work done by the electric field is exactly the opposite: W_field = -q(V_B - V_A) = q(V_A - V_B). NEET usually asks for the work done to move the charge (external), which is q(V_B - V_A).
If a positive charge moves to higher potential, is the work positive or negative?
Positive. For a positive charge, moving to higher potential means V_B > V_A, so W = q(V_B - V_A) is positive: you must push against the field and do positive work. Moving a positive charge to lower potential gives negative external work (the field helps). Flip the sign of the whole result if the charge is negative.
⚠️ The NEET trap ✗ Reading a diagram where the equipotential lines are arranged differently in each figure and concluding the work done is different in each case. ✓ Work depends only on the potentials at the two endpoints A and B. If A and B lie on the same two equipotential values (e.g. 10 V to 40 V) in every figure, then delta V = 30 V and W = q(30) is identical in all figures, regardless of spacing or arrangement. 🧠 Do not judge work by how the lines look on paper. Only the start and end potentials count.
Real NEET questions
NEET 2017
The diagrams (a), (b), (c), (d) show regions of equipotentials (with values 10 V, 20 V, 30 V, 40 V arranged differently). A positive charge is moved from A to B in each diagram. Which statement is correct?
A · Maximum work is required to move q in figure (c)
B · In all the four cases the work done is the same ✓
C · Minimum work is required to move q in figure (a)
D · Maximum work is required to move q in figure (b)
Solution: Step 1: Write the formula. Work to move charge q from A to B is W = q(V_B - V_A). It depends ONLY on the endpoint potentials. Step 2: In every figure, point A sits on the 10 V equipotential and point B on the 40 V equipotential. Step 3: So delta V = V_B - V_A = 40 - 10 = 30 V in all four figures. Step 4: Therefore W = q(30 V) is the same in every case, no matter how the lines are drawn. The arrangement or spacing does not change the work. Correct answer: (B).
ReNEET 2026
A unit positive point charge is taken slowly through a thin tube inside a charged dielectric sphere of radius R with uniform positive charge density rho. Points A and B are at distances 2R and 3R from the centre. The magnitude of total work done is (rho R^2)/(n epsilon_0). Find n.
A · 2
B · 6
C · 9
D · 18 ✓
Solution: Step 1: Both A (2R) and B (3R) lie OUTSIDE the sphere, so the sphere acts like a point charge Q = rho x (4/3)pi R^3 at its centre. Step 2: Outside potential is V = kQ/r. So V_A = kQ/(2R) and V_B = kQ/(3R). Step 3: Work W = q(V_B - V_A) with q = 1. W = kQ[1/(3R) - 1/(2R)] = kQ(-1/6R). Step 4: Magnitude |W| = kQ/(6R). Substitute k = 1/(4 pi epsilon_0) and Q = rho(4/3)pi R^3: |W| = [1/(4 pi epsilon_0)] x [1/(6R)] x rho(4/3)pi R^3 = (rho R^2)/(18 epsilon_0). Step 5: So n = 18. Correct answer: (D).
Solved Electrostatic Potential And Capacitance NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for work done in moving a charge between two points?
W = q(V_B - V_A), where q is the charge, V_A is the potential at the starting point and V_B at the ending point. It gives the external work needed to move the charge slowly.
Why is work done zero on an equipotential surface?
Because every point on an equipotential surface has the same potential, so V_B - V_A = 0, making W = q(0) = 0 no matter the path.
Does work done depend on the path?
No. The electrostatic force is conservative, so work done depends only on the initial and final potentials, not on the path taken between them.
Is work done by the field the same as work done by me?
No, they are opposite in sign. Work by external agent = q(V_B - V_A); work by the field = -q(V_B - V_A).
How do I find work done bringing a charge from infinity?
At infinity V = 0, so W = q(V_B - 0) = qV_B, where V_B is the potential at the final point.
