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)

Electrostatics: JEE Main 2023 Physics question

JEE Main 2023 · 8 Apr, Evening shift · Q7022

JEE Main 20238 Apr, Evening shiftPhysicsMediumElectrostatics167× since 2002

Electric potential at a point '

\mathrm{P}

' due to a point charge of

5 \times 10^{-9} \mathrm{C}

50 \mathrm{~V}

. The distance of '

\mathrm{P}

' from the point charge is: (Assume,

\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{+9} ~\mathrm{Nm}^{2} \mathrm{C}^{-2}

)

JEE Main 2023 Electrostatics Question with Solution (8 Apr Evening shift, Q7022) | JEE Arc