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? 


1
Expert's answer
2022-02-28T13:52:20-0500

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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