There are two identical, positively charged conducting spheres fixed in space. T

Question

There are two identical, positively charged conducting spheres fixed in space. The spheres are 47.6 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0795 N. Then, a thin conducting wire connects the spheres, redistributing the charge on each sphere. When the wire is removed the spheres still repel but with a force of F2 = 0.115 N. Using this information, find the initial charge on each sphere, q1 and q2 if initially q1< q2. The Coulomb Gorce Constant is 8.99x10^9 N*M^2/C^2

q1 and q2 should be in coulumbs. Please go through step by step, so I can understand.

Explanation / Answer

q1 and q2 are the charges. As,the spheres are identical they will have same capacitance and according to the rule V = q/C the will distribute the charges equally(As,then only the potential will be same).

So, from 2nd condition we get, q^2*k/(.476)^2 =0.115N => q= 0.17*10^-5 C

Now,the total charge was 0.34*10^-5C.

So, let the charge q2 be x C,then q1=(0.34*10^-5-x) C.

From, first condition we get 8.99*10^9(0.34*10^-5*x-x^2)/(.476)^2 = 0.0795

=>-x^2+0.34*10^-5*x = 0.2*10^-11

=>x= {0.34*10^-5 (+-)sqrt(0.356*10^-11)}/2 = {0.34*10^-5 (+-)0.19*10^-5}/2

So,q1=0.75*10^-6 C,q2 =2.65*10^-6 C(Ans.)

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