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11111nata11111 [884]

1 year 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?

1 answer:

erma4kov [3.2K]1 year 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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The question is incomplete. Here is the entire question.

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