Answer to Question #173792 in Physics for Isey

Question #173792

If the source charge is +4.31x10^3 C and the test charge is placed 7.11x10^-12m away from it?

Expert's answer

... then the electric field is

"E=k\\frac{q}{r^2}=9\\cdot10^9\\cdot\\frac{4.31\\cdot10^{3}}{(7.11\\cdot10^{-12})^2}=7.67\\cdot10^{35}\\ (V\/m)" . Answer

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