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 34.8 cm apart (center to center) and repel each other with an electrostatic force of F1 = 0.0630 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. The Coulomb Force constant is k=8.99*10^9

Solve for q1 and q2.

Map Sapling Learning There are two identical, positively charged conducting spheres fixed in space. The spheres are 34.8 cm apart (center to center) and repel each other with an electrostatic force of F1 0.0630 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 formation, find the initial charge on each sphere, q1 and q2 if initially qh q2. The Coulomb Force constant is k 1/(4meo) 8.99 x 109 N. m2/c2 Number Number 92

Explanation / Answer

Initially:

F1 = kq1q2/d²

0.0630 = 8.99 x 109 q1q2/0.3482

q1q2 = 8.4867 x 10-13

q1 = 8.4867 x 10-13 /q2

Since the final charges will be equal, say Q on each sphere:

F2 = kQ.Q/d²

F2 = kQ²/d²

0.100 = 8.99 x 109 Q2/0.3482

Q = 1.160645 x 10-6

The total charge is 2Q (because Q is on each sphere)

Since the charge has been redistributed (none gained or lost overall)

2Q = q1 + q2

2Q = 8.4867 x 10-13/q2 + q2

2.32129 x 10-6q2 = 8.4867 x 10-13 + q2²

q2 = 4.54 x 10-7 C

q1 = 1.86 x 10-6 C

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