Electric Potential at a Point — V = kq/r

easy CBSE JEE-MAIN JEE-ADVANCED NCERT Class 12 4 min read

Question

A point charge of q = +2 µC is placed in free space. Find the electric potential at a point P located r = 0.3 m away from the charge. Also, find the work done in bringing a test charge of q₀ = +1 µC from infinity to point P.


Solution — Step by Step

Electric potential due to a point charge is:

V=kqrV = \frac{kq}{r}

Here, k = 9 × 10⁹ N·m²/C², q is the source charge, and r is the distance from the charge to the point. Unlike electric field, potential is a scalar — no direction needed.

V=9×109×2×1060.3V = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{0.3} V=180000.3=60000 VV = \frac{18000}{0.3} = 60000 \text{ V}

So V = 6 × 10⁴ V at point P.

The relation between work done and potential is:

W=q0VW = q_0 \cdot V

We use this because potential at infinity is zero, so the potential difference is simply V itself.

W=1×106×6×104=0.06 JW = 1 \times 10^{-6} \times 6 \times 10^4 = 0.06 \text{ J}

W = 0.06 J (or 6 × 10⁻² J).

Both charges are positive. Bringing a positive test charge toward another positive charge requires work done against the repulsive force — so W is positive. The answer checks out physically.


Why This Works

Electric potential at a point is defined as the work done per unit positive charge in bringing a test charge from infinity to that point, with no acceleration (quasi-static process). The formula V = kq/r directly gives us this quantity for a point charge.

The key reason potential varies as 1/r (while electric field varies as 1/r²) is that potential is obtained by integrating the electric field over distance. Each integration step picks up one factor of r in the denominator.

Since potential is a scalar, when multiple charges are present, we simply add the potentials algebraically — no vector addition needed. This is what makes potential extremely useful for problems with complex charge distributions.

For JEE, the scalar nature of potential is a high-weightage concept. Superposition of potentials (just add numbers) versus superposition of electric fields (vector addition) — this distinction has appeared in multiple PYQs.


Alternative Method — Using Energy Directly

Instead of finding V first, we can directly use the definition of potential energy:

U=kqq0rU = \frac{kq \cdot q_0}{r} U=9×109×2×106×1×1060.3U = \frac{9 \times 10^9 \times 2 \times 10^{-6} \times 1 \times 10^{-6}}{0.3} U=18×1030.3=0.06 JU = \frac{18 \times 10^{-3}}{0.3} = 0.06 \text{ J}

Since both charges start at infinity (U = 0) and end with U = 0.06 J, the work done equals this potential energy: W = 0.06 J. Same answer, different route.

This approach is faster when the question directly asks for work done or potential energy between two specific charges.


Common Mistake

Students often confuse electric potential V = kq/r with electric field E = kq/r². The field has in the denominator and is a vector. The potential has r (not squared) and is a scalar. In CBSE board exams, writing V = kq/r² is one of the most common errors and costs full marks. Remember: potential comes from integrating the field, which “reduces” the power of r by 1.

A second trap: when the source charge is negative, the potential is negative. A negative potential means we actually get energy back when bringing a positive test charge from infinity — the charges attract. Always carry the sign of q through the calculation.

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