Question

A point charge Q creates an electric potential of +130 V at a distance of 25 cm . What is Q?

Answer 1

If a charge is placed at a point, it will be having some electric potential, which can be calculated using

$V=\frac{kQ}{r}$

where k is electrostatic constant, Q is the charge and r is the distance at which potential due to charge is to be calculated,

$k=\frac{1}{4\pi \varepsilon _0}=9*10^9Nm^2/C^2$

where $\varepsilon _0$  is the absolute electrical permittivity and $\varepsilon _0=8.85*10^{-12}C^2N^{-1}m^{-2}$

given, V = +130V, r = 25 cm = 0.25 m

so using above relation,

$+130=\frac{9*10^9*Q}{0.25}$

$Q=\frac{+130*0.25}{9*10^9}C$

$Q=+3.61*10^{-9}C$

$Q=+3.61nC$