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 41.2 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0765 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 force constant is k = 8.99 x 109 Nm2/C2.

Explanation / Answer

Let the charges on spheres be q1 and q2 respectively.

So, Force of attraction, F = k*q1*q2/r^2 = 0.0765

'where r = distance between them = 41.2 cm = 0.412 m

k = 9*10^9

So, q1*q2 = 0.0765*0.412^2/(9*10^9) = 1.44*10^-12

When they are connected, the charges rearrange such that both have equal charges on them

Let the charge now on both be q and thus total charge on both = 2*q

So, by conservation of charge, q1+q2 = 2*q

Now, new Force of repulsion, F' = k*q*q/r^2 = 0.115

So, q^2 = F'*r^2/k = 0.115*0.412^2/(9*10^9) = 2.17*10^-12

So, q = 1.47*10^-6

So, q1+q2 = 2*q = 2.95*10^-6

So, q2 = 2.95*10^-6 - q1

But, q1*q2 = 1.44*10^-12

So, q1*(2.95*10^-6 - q1) = 1.44*10^-12

So, q1 = 0.62*10^-6 C <------------answer

q2 = 2.33*10^-6 C <----------answer

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects