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A solid conducting sphere of radius R1 = 1.790 m𝑅1 = 1.790 m has a charge of Q = 6.749 μC distributed evenly on its surface. A second solid conducting sphere of radius R2 = 0.6425 m𝑅2 = 0.6425 m is initially uncharged. The second sphere is located a distance of 10 m from the first sphere. The two spheres are momentarily connected with a wire and then the wire is disconnected. What is the charge on the second sphere? (You may enter your calculation using scientific notation.)  C

Question

A solid conducting sphere of radius R1 = 1.790 m𝑅1 = 1.790 m has a charge of Q = 6.749 μC distributed evenly on its surface. A second solid conducting sphere of radius R2 = 0.6425 m𝑅2 = 0.6425 m is initially uncharged. The second sphere is located a distance of 10 m from the first sphere. The two spheres are momentarily connected with a wire and then the wire is disconnected. What is the charge on the second sphere? (You may enter your calculation using scientific notation.)  C

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Solution 1

The charge will distribute evenly between the two spheres when they are connected by a wire. The amount of charge on each sphere after disconnection will be proportional to the radius of each sphere.

The total charge (Q_total) is the sum of the charges on both spheres, which is 6.749 μC in this case.

The charge on the second sphere (Q2) can be calculated using the formula:

Q2 = Q_total * (R2 / (R1 + R2))

Substituting the given values:

Q2 = 6.749 μC * (0.6425 m / (1.790 m + 0.6425 m))

After calculating the above expression, we get the charge on the second sphere.

This problem has been solved

Solution 2

The charge will distribute evenly between the two spheres when they are connected by a wire. The amount of charge on each sphere after disconnection will be proportional to the radius of each sphere.

The total charge (Q_total) is the sum of the charges on both spheres, which is 6.749 μC in this case.

The charge on the second sphere (Q2) can be calculated using the formula:

Q2 = Q_total * (R2 / (R1 + R2))

Substituting the given values:

Q2 = 6.749 μC * (0.6425 m / (1.790 m + 0.6425 m))

After calculating the above expression, we get the charge on the second sphere.

This problem has been solved

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