Question

The electric field 0.500 m from acharge is 3370 N/C, pointing awayfrom the charge. What is thecharge? Include the sign of thecharge, + or -(The answer is *10^-8 C. Just fillin the number, not the power.)(Unit = C)

Answer 1

To solve this problem, we have to use the electric field definition.

$E=K\cdot\frac{q}{r^2}$

Replacing the needed information, we have

$3370=9\times10^9\cdot\frac{q}{(0.5)^2}$

Let's solve for q

$\begin{gathered} q=\frac{3370\cdot0.25}{9\times10^9} \\ q=93.6\times10^{-9}C \\ q=9.36\times10^{-8}C \end{gathered}$

Therefore, the answer is 9.36*10^-8 C.