Question #174846

calculate the electric Field intensity at a point 15 cm from a charge of 10× 10^ -3C


1
Expert's answer
2021-03-24T19:40:23-0400

By the definition of the electric field intensity, we have:


E=kQr2=9109 Nm2C210103 C(0.15 m)2=4.0109 NC.E=\dfrac{kQ}{r^2}=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot10\cdot10^{-3}\ C}{(0.15\ m)^2}=4.0\cdot10^9\ \dfrac{N}{C}.

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