Physics · Electrostatic Potential And Capacitance · NEET
Potential is V = kQ/r (one r in the bottom). Electric field is E = kQ/r² (two r's, r squared). Reason: field is a force-like quantity that falls as 1/r², and potential is that field added up (integrated) over distance from infinity, so it loses one power of r and becomes 1/r. Quick check: E has units N/C, V has units volts = N·m/C, so V has one extra metre, meaning one less r in the denominator.
Potential is defined as the work done per unit positive charge to move it from a reference point to point P. We pick infinity as the reference because at infinity the point charge's field is zero, so V(infinity) = 0. Starting from a clean zero makes the maths simple. So V at P = work done from infinity to P, divided by the test charge.
No. We divide the work by the test charge (V = W/q₀), so the test charge cancels out. Potential is a property of the source charge Q and the distance r only. That is why we use a unit positive test charge in the derivation.
It is a scalar. It has magnitude and a sign (+ or −) but no direction. For many charges you just add the values V = kQ₁/r₁ + kQ₂/r₂ + ... with their signs, no vector triangle needed. This makes potential much easier to add than electric field.
Work done by external agent per unit charge = −∫E·dr from infinity to r. With E = kQ/r² pointing outward, V = −∫(from ∞ to r) kQ/r² dr = kQ[1/r] evaluated from ∞ to r = kQ(1/r − 0) = kQ/r. The 1/r² inside the integral becomes 1/r after integrating.
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.
It is V = kQ/r, the work done to bring a unit positive charge from infinity to a point at distance r from the charge Q, where k = 9 × 10⁹ N·m²/C².
Potential is the electric field integrated over distance. Integrating 1/r² with respect to r gives 1/r, so potential falls one power slower than the field.
For +Q the potential is positive (V = +kQ/r); for −Q it is negative (V = −kQ/r). Distance r is always positive, so only the charge's sign decides the sign of V.
Zero. As r tends to infinity, V = kQ/r tends to 0. That is why infinity is used as the reference point in the derivation.
No, it is a scalar. It has a value and a sign but no direction, so potentials from many charges are added by simple algebra.
