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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)

Respuesta :

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

[tex]E=K\cdot\frac{q}{r^2}[/tex]

Replacing the needed information, we have

[tex]3370=9\times10^9\cdot\frac{q}{(0.5)^2}[/tex]

Let's solve for q

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

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

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