Physics · Electrostatic Potential And Capacitance · NEET
The minus sign shows direction. Work done by the field on a positive charge is positive when it moves to lower potential, so the field pushes charges from high V to low V. Since dV/dr is the rate at which potential increases with distance, the field points the opposite way - toward decreasing potential. That is why E = -dV/dr. The minus sign is not about V being negative; it only fixes the direction of E.
No. E depends on the slope dV/dr, not on the value of V. You can have V = 0 at a point while V is still changing around it, so dV/dr is not zero and E is not zero. A common example is the midpoint between +q and -q (equatorial region of a dipole): V = 0 there but E is not zero. So never conclude E = 0 just because V = 0.
dV/dr is the true slope of V at that exact point (a derivative). V/r is just the average value of potential divided by distance and is only correct in special cases. For a point charge, V = kQ/r, so dV/dr = -kQ/r^2, giving E = kQ/r^2. Note E = V/r would give kQ/r^2 too here by coincidence, but for other configurations V/r is wrong. Always use the slope dV/dr, not V/r, unless the field is uniform.
The field magnitude equals the negative slope of the V-r graph: E = -(slope). A steep graph means a strong field; a flat region (constant V) means E = 0, which is what happens inside a conductor or on an equipotential surface. If V rises with r, the slope is positive so E points toward smaller r; if V falls with r, E points toward larger r.
Potential gradient is dV/dr, the change in potential per unit distance. Its SI unit is volt per metre (V/m), which is exactly the same as newton per coulomb (N/C), the unit of electric field. This is why E and V/m are interchangeable units - they come straight from E = -dV/dr.
If a conducting sphere of radius R is charged, then the electric field at a distance r (r > R) from the centre of the sphere would be (V = potential on the surface of the sphere):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The electric field equals the negative gradient of potential: E = -dV/dr. The field points toward decreasing potential, and its magnitude equals the rate at which potential changes with distance.
The unit of potential gradient dV/dr is volt per metre (V/m), which is the same as newton per coulomb (N/C), the unit of electric field.
Yes. Inside a charged conductor and everywhere on an equipotential surface, V is constant (not necessarily zero) but its slope is zero, so E = 0. Constant potential always means zero field.
E = V/d (or V/r) is only the special case of a uniform field, where V changes at a constant rate. The general relation is E = -dV/dr, using the actual slope of V at that point.
A positive charge naturally moves toward lower potential energy, so the field that pushes it must point from high V to low V. The minus sign in E = -dV/dr encodes exactly this direction.
