Question

There are two identical, positively charged conducting spheres fixed in space. The spheres are 43.0 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0675 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 = 0.100 N. Using this information, find the initial charge on each sphere, q1 and q2 if initially q1 < q2. Coulomb force constant, k, = 8.99E9

q1=?

q2=?

Answers in C

Answer 1

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There are two identical, positively charged conducting spheres fixed in space. The spheres are 44.0 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 = 0.100 N. Using this information, find the initial charge on each sphere, q1 and q2 if initially q1

kq1q2/r^2=0.0765

kq^2/r^2=0.1

8.99x10^9*q^2/0.44^2=0.1

So q=1.467 x10^-6 C

q=(q1+q2)/2

So q1+q2=2q=2.935 x10^-6 C

kq1q2/r^2=0.0765

8.99x10^9*q1*q2/0.44^2=0.0765

So q1q2=1.647 x 10^-12

Solving for q1 and q2 gives

q1=0.756 x10^-6 C

q2=2.18 x10^-6 C