Answer to Question #302137 in Physics for Achi

Question #302137

The electric field has a magnitude of 3.0 N/C at a distance of 30 cm from a point charge. What is the charge?

Expert's answer

The electric field due point charge

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

Hence, the charge

"q=Er^2\/k\\\\\n=3.0*0.3^2\/9*10^9=3*10^{-11}\\:\\rm C"

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