A charge of 3µC is used to test the electric field of a central charge of 6C that causes a force of 800N. What is the magnitude of the electric field?
F=qE→E=F/q=800/(3⋅10−6)=267⋅106 (V/m)F=qE\to E=F/q=800/(3\cdot10^{-6})=267\cdot10^6\ (V/m)F=qE→E=F/q=800/(3⋅10−6)=267⋅106 (V/m) . Answer
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