Question

What magnitude charge creates a 15.43 N/C electric field at a point 2.53 m away?

Answer 1

Given:

The electric field is E = 15.43 N/C

The distance is r = 2.53 m away.

Required: Magnitude of charge.

Explanation:

The magnitude of charge can be calculated by the formula

$\begin{gathered} E=\frac{kQ}{r^2} \\ Q=\frac{Er^2}{k} \end{gathered}$

Here, Coulomb's constant is

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

On substituting the values, the magnitude of charge will be

$\begin{gathered} Q=\frac{15.43\times(2.53)^2}{9\times10^9} \\ =1.097\times10^{-8}\text{ C} \end{gathered}$

Final Answer: The magnitude of the charge is 1.097 x10^(-8) C