sandyyknguyen1391 sandyyknguyen1391

12-03-2020
Physics

contestada

Two positive charges of magnitude 20 micro C and 100 micro C are placed at a distance of 150cm. Calculate the electric force of repulsion between them.

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iyohalucky7 iyohalucky7

12-03-2020

Answer:800N

Explanation:

Force of repulsion = kq1q2/r^2

Q1= 20*10^-6

Q2=100*10^-6

R=150cm=0.15m

K= 9*10^9

F= 2*10^-5*1*10^-4*9*10^9/(0.15)^2

F= 2*1*9*10^-5-4+9/(0.15)^2

F= 18*10^0/(0.15)^2

F=18/0.0225

F=800N

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