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}*10^(-3)\text{ C} \end{gathered}`

The value of the constant is

`k=\text{ 8.99}*10^9\text{ N m}(^2)/(C^2)`

The electric potential is

`V=\text{ 4.2}*10^4\text{ V}`

To find the distance.

Explanation:

The distance can be calculated by the formula

`\begin{gathered} V=(kQ)/(r) \\ r=(kQ)/(V) \end{gathered}`

On substituting the values, the distance will be

`\begin{gathered} r=(8.99*10^9*8*10^(-3))/(4.2*10^4) \\ =\text{ 1712.38 m} \end{gathered}`

Thus, the distance is 1712.38 m.