Question

At what distance from a point charge of 8.0 mC would the electric potential be 4.2 x 10^4 V? (kc = 8.99 x 10^9 N·m^2/C^2)

Answer 1

Given:

The charge is

$\begin{gathered} Q=\text{ 8 mC} \\ =\text{ 8}\times10^{-3}\text{ C} \end{gathered}$

The value of the constant is

$k=\text{ 8.99}\times10^9\text{ N m}\frac{^2}{C^2}$

The electric potential is

$V=\text{ 4.2}\times10^4\text{ V}$

To find the distance.

Explanation:

The distance can be calculated by the formula

$\begin{gathered} V=\frac{kQ}{r} \\ r=\frac{kQ}{V} \end{gathered}$

On substituting the values, the distance will be

$\begin{gathered} r=\frac{8.99\times10^9\times8\times10^{-3}}{4.2\times10^4} \\ =\text{ 1712.38 m} \end{gathered}$

Thus, the distance is 1712.38 m.