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SashulF [63]
2 years ago
6

Two positive charges ( 8.0 mC and 2.0 mC) are separated by 300 m. A third charge is placed at distance r from the 8.0 mC charge

in such a way that the resultant electric force on the third charge due to the other two charges is zero. The distance r is
Physics
1 answer:
WINSTONCH [101]2 years ago
8 0

Answer:

r=200m

Explanation:

From the question we are told that:

Charges:

Q_1=8.0mC

Q_2=2.0mC

Q_3=8.mC

Distance d=300m

Generally the equation for Force is mathematically given by

F=\frac{kq_1q_2}{r^2}

Therefore

F_{32}=F_{31}

\frac{q_2}{(300-r)^2}=\frac{q_1}{r^2}

\frac{2*10^{-3}}{(300-r)^2}=\frac{8*10^{-3}}{r^2}

r=2(300-r)

r=200m

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