Two positive charge distant 0.1 m apart,repel each other with a force of 18 N.If the sum of the charges be 9 micro coulomb,then calculate their seperate forces
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Hello Shahrukh, let one charge be q micro coulomb, then the other will be 9-q micro coulomb.
By Coulomb's law, the force F = 9*10^9*q1*q2/r^2
r=0.1m and q1=q*10^-6C and q2 = (9-q)*10^-6 C
Plugging the values,
18 = 9*10^9*q(9-q)*10^-12 / 0.01
Simplifying we get q^2 -9q+20=0
Factorizing (q-4)*(q-5)=0
Hence the charges have to be 4uC and 5uC. u stands for mu symbol, ie micro which equals to 10^-6
By Coulomb's law, the force F = 9*10^9*q1*q2/r^2
r=0.1m and q1=q*10^-6C and q2 = (9-q)*10^-6 C
Plugging the values,
18 = 9*10^9*q(9-q)*10^-12 / 0.01
Simplifying we get q^2 -9q+20=0
Factorizing (q-4)*(q-5)=0
Hence the charges have to be 4uC and 5uC. u stands for mu symbol, ie micro which equals to 10^-6