PhysicsMedium46×since 2002Q7022Electric potential at a point 'P\mathrm{P}P' due to a point charge of 5×10−9C5 \times 10^{-9} \mathrm{C}5×10−9C is 50 V50 \mathrm{~V}50 V. The distance of 'P\mathrm{P}P' from the point charge is: (Assume, 14πε0=9×10+9 Nm2C−2\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{+9} ~\mathrm{Nm}^{2} \mathrm{C}^{-2}4πε01=9×10+9 Nm2C−2 )A0.9 cmB90 cmC3 cmD9 cmCheck answerSkip