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11111nata11111 [884]
4 years ago
8

Two identical conducting spheres, A and B, sit atop insulating stands. When they are touched, 1.51 × 1013 electrons flow from sp

here B to sphere A. If the total net charge on the spheres is +3.68 μC, what was the initial charge on sphere B?
Physics
1 answer:
erma4kov [3.2K]4 years ago
6 0

Answer:

A = -0.576 μC

B = 4.256 μC

Explanation:

Suppose a single electron charge is 1.6\times10^{-19}C. Then the total charge that is flowing from B to A is:

1.6\times10^{-19} * 1.51 \times 10^{13} = 2.416\times10^{-6}C = 2.416 \mu C

Let A and B be the initial charge of spheres A and B, respectively. Since the net charge is 3.68μC we have the following equation

A + B = 3.68 (1)

When they touch 2.416μC flows from B to A, then they are equal, so we have the following equation

A + 2.416 = B - 2.416

-A + B = 2.416 + 2.416 = 4.832 (2)

Add equation (1) to equation (2) we have

2B = 3.68 + 4.832 = 8.512

B = 8.512 / 2 = 4.256 \mu C

A = 3.68 - B = 3.68 - 4.256 = -0.576 \mu C

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