Question

A conducting sphere of radius r1= 0.38m has a total charge of Q=.75 uC. A second uncharged conducting sphere of radius r2= 0.28m is then connected to the first by a thin conducting wire. The spheres are separated by a very large distance to their size. What is the total charge on sphere two, Q2, in C?

Answer 1

Answer:

1.02 uC

Explanation:

Two charged conducting spheres will almost always attract each other at close approach, even when they have like charges. Like charges repel each other, opposite charges attract but this is not the case for conducting spheres. If two spheres with like charges are brought close enough, they will attract.

From the question, the parameters given are; radius r1= 0.38m, total charge of Q=.75 uC, A second uncharged conducting sphere of radius r2= 0.28, total charge on sphere two, Q2=??.

r1×Q1= r2× Q2

0.38 m× 0.75 uC= Q2 × 0.28 m

Q2= 1.02 uC

Answer 2

Answer:

The Q₂ is 0.318 μC

Explanation:

The charge flows is the same on both, then:

$V_{1} =\frac{kQ_{1} }{r_{1} } \\V_{2} =\frac{kQ_{2} }{r_{2} } \\Q_{2} =Q-Q_{1} \\\frac{kQ_{1} }{r_{1} } =\frac{k*(Q-Q_{1}) }{r_{2} } \\Q_{1} =\frac{\frac{Q}{r_{2} } }{(1/r_{1})+(1/r_{2}) }$

But:

$\frac{1}{r_{1} } +\frac{1}{r_{2} }=\frac{1}{0.38} +\frac{1}{0.28} =6.2$

Q = = 0.75 μC

Replacing:

$Q_{1} =\frac{\frac{0.75}{0.28} }{6.2} =0.432\mu C$

The Q₂ is equal:

Q₂ = 0.75 - 0.432 = 0.318 μC