JEE MainPhysicsElectrostatics
At a certain point in space, the electric potential due to a point charge is 36 V and the magnitude of the electric field due to the same charge is 12 V/m . The magnitude of the point charge is: (Assume 1 4 ₀ = 9 10^9 N m ^2 C ⁻² )
Options
- A4 3 nC
- B4 3 nC
- C1 3 nC
- D12 nC
Correct answer
D. 12 nC
Step-by-step solution
The electric potential V and the electric field magnitude E due to a point charge Q at a distance r are given by: V = 1 4 ₀ Q r E = 1 4 ₀ Q r^2 Dividing the expression for potential by the expression for electric field gives the distance r : V E = 1 4 ₀ Q r 1 4 ₀ Q r^2 = r Substituting the given values: r = 36 12 = 3 m Now, substitute the value of r back into the potential equation to find Q : V = 1 4 ₀ Q r 36 = (9 10^9) Q 3 36 = 3 10^9 Q Q = 36 3 10^9 = 12 10⁻⁹ C Q = 12 nC Answer: 12 nC