Electric potential at a point 'P' due to a point charge of 5 × 10⁻⁹ C is 50 V. The distance of 'P' from the point charge is:
(Assume, 1/(4 π ε₀)=9 × 10⁺⁹ Nm² C⁻² )

Solution

Physics sir · 3 steps

Potential falls as 1/r, not 1/r², so rearrange V = kq/r for r and watch the units.

1. Point-charge potential. At a distance r from a point charge q, the potential is V = kq/r, with k = 1/4πε₀. Rearranged, this gives r directly.

   r = kq/V = ((9 × 10⁹) × (5 × 10⁻⁹))/50

2. Put in numbers. The powers of ten cancel: 9 × 10⁹ × 5 × 10⁻⁹ = 45. Then r = 45/50 = 0.9 m, because SI inputs give the answer in metres.
3. So. r = 0.9 m = 90 cm, which is option (b).

The idea to keep. With all values in SI units, r = kq/V comes out in metres. Convert to centimetres only at the end, by multiplying by 100.

Answer from the NTA final answer key. Solution by PrepWiser.