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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

User Spender
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1 Answer

25 votes
25 votes

Answer:


r=200m

Step-by-step 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=(kq_1q_2)/(r^2)

Therefore


F_(32)=F_(31)


(q_2)/((300-r)^2)=(q_1)/(r^2)


(2*10^(-3))/((300-r)^2)=(8*10^(-3))/(r^2)


r=2(300-r)


r=200m

User Chad DeShon
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