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 35.2 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 0.115 N. Using this information, find the initial charge on each sphere, q1 and q2 if initially q1 q2. The Coulomb Force constant is k = 114TTEo) = 8.99 × 109 N·m2/C2 Number 41 Number 42

Explanation / Answer

Initially force will be given by:

F = kq1q2/r^2

Using given values:

0.0675 = 9*10^9*q1*q2/0.352^2

q1*q2 = 0.0675*0.352^2/(9*10^9) = 9.292*10^-13 C

After redistributin charge on each sphere will be same,

q = (q1+q2)/2

Now force will be given by:

F1 = k*q*q/r^2

q = sqrt (F1*r^2/k)

q = sqrt (0.115*0.352^2/(9*10^9))

q = 1.258*10^-6

So,

(q1 + q2)/2 = 1.258*10^-6 C

q1 + q2 = 2.516*10^-6 C

q1*q2 = 9.292*10^-13 C

Solving, both equation

q2 = 2.06631*10^-6 , AND q1 = 0.44969*10^-6

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