Question

What magnitude charge creates a 19.59 N/C electric field at a point 9.52 m away?

Answer 1

Given:

The magnitude of the electric field is: E = 19.59 N/C

The distance at which the electric field is measured is: r = 9.52 m

To find:

The magnitude of the charge.

Explanation:

The expression for the electric field intensity E is given as:

$E=k\frac{q}{r^2}$

Here, k = 4πε₀ ≈ 8.99 × 10⁹ N.m²/C², q = the magnitude of the charge.

Rearranging the above equation, we get:

$q=\frac{Er^2}{k}$

Substituting the values in the above equation, we get:

$\begin{gathered} q=\frac{19.59\text{ N/C}\times\text{ \lparen}9.52\text{ m\rparen}^2}{8.99\times10^9\text{ N.m}^2\text{/C}^2} \\ \\ q=\frac{19.59\text{ N/C}\times90.6304\text{ m}^2}{8.99\times10^9\text{ N.m}^2\text{/C}^2} \\ \\ q=1.975\times10^{-7}\text{ C} \\ \\ q=0.1975\text{ }\times10^{-6}\text{ C} \\ \\ q\approx0.1975\text{ }\mu\text{C} \end{gathered}$

Final answer:

The magnitude of the charge is 0.1975 microCoulombs.