Answer to Question #299320 in Physics for yumi

Question #299320

Calculate the magnitude and direction of the electric field at a point P which is 30 cm to the right of a point charge Q = -3.0 x 10-6 C


1
Expert's answer
2022-02-18T08:21:57-0500

Given:

Q=3.0106CQ = -3.0 *10^{-6}\:\rm C

r=0.3mr=0.3\:\rm m


The magnitude of electric field at the point P

E=kQr2=91093.01060.32=3105V/mE=k\frac{|Q|}{r^2}=9*10^9*\frac{3.0 *10^{-6}}{0.3^2}=3*10^5\:\rm V/m

The field is directed toward the charge.


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