Physics · Electrostatic Potential And Capacitance · NEET
It is a curve, not a straight line. Since V = kQ/r, V is inversely proportional to r. As r doubles, V becomes half; as r triples, V becomes one-third. This gives a smooth downward-bending curve (a rectangular hyperbola shape), steep near the charge and flat far away. A straight line would mean V is constant or changes by a fixed amount per metre, which is wrong here.
Field is E = kQ/r² and potential is V = kQ/r. Potential is the work per unit charge to bring a charge from infinity to that point, and integrating the 1/r² field over distance gives a 1/r result. So V drops more slowly than E. Near the charge E rises much faster than V; far away both go toward zero, but E dies out quicker.
No. As r goes to infinity, V = kQ/r approaches zero but never actually reaches it. The r-axis (V = 0) is a horizontal asymptote. Also the curve never crosses to the other side of the V-axis, because r cannot be zero or negative for a real point outside the charge.
For +q, V is positive everywhere, so the curve sits above the r-axis and falls toward zero from above. For -q, V = -kq/r is negative everywhere, so the curve is the mirror image below the r-axis, rising toward zero from below. The shape (1/r) is the same; only the sign flips.
At r = 0 the formula V = kQ/r gives division by zero, so V tends to infinity. Physically this is because a true point charge has all its charge at a single point. For real objects like a charged sphere this never happens, because you cannot get closer than the surface, and inside a conductor the potential stays constant.
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.
V = kQ/r, where k = 1/(4πε₀) = 9 × 10⁹ N m²/C². Potential V is inversely proportional to distance r, giving a 1/r curve.
Yes, in shape. Because V ∝ 1/r, the curve is a rectangular hyperbola branch: steep near the charge, flattening toward the r-axis at large distances without touching it.
The slope dV/dr is negative and its magnitude gives the electric field, since E = -dV/dr. Where the curve is steep (near the charge), the field is large; where it is flat (far away), the field is small.
For -q the potential is negative everywhere, so the whole curve is reflected below the r-axis. It rises toward zero from below as r increases, but keeps the same 1/r shape.
NEET often shows a graph and asks whether it is V vs r or E vs r, or asks about the sign for +q vs -q. Knowing V ∝ 1/r (gentle) versus E ∝ 1/r² (steep) helps you pick the correct curve quickly.
