Question

If two point chargeg are kept at distance of 3m find maximum force between tham​

Answer 1

Answer:

q1=charge1

q2=charge 2

maximum force= 9*10^9* q1*q2/ 3^2

=10^9*q1*q2

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Answer 2

Two point charges q1 and q2 are 3m apart, and their combined charge is 20 micro C. If one repels the other with a force of .075 N, what are the two charges?

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Here, q1+q2=20x10^-6C…………….(1)

The force of repulsion, according to Coulomb's law,

F=kq1q2/r^2=kq1(20x10^-6 - q1)/(3)^2 or

75x10^-3=9x10^9 q1(20x10^-6- q1)/9=10^9q1(20x10^-6 -q1). Therefore ,

75x10^-12=q1(20x10^-6- q1). Therefore,

75x10^-12= 20x10^-6 q1-q1^2

Or

q1^2–20x10^-6 q1 +75x10^-12 =0……….(2)

This is quadratic equation in q1. The roots of this equation are

q1=15x10^-6C and then from equation (1) q2=5x10^-6C

and

q1=5x10^-6C and q2=15x10^-6C.