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Anettt [7]
3 years ago
9

Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with the

ir diameters. The magnitude of the electrostatic force acting on sphere 2 due to sphere 1 is F = 0.42 N. Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere , then to sphere 2, and finally removed. What is the magnitude of the electrostatic force F' that now acts on sphere 2?
Physics
1 answer:
scoundrel [369]3 years ago
3 0

Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = \frac{kQ_{1}Q_{2} }{r^{2} }

F= \frac{K(Q/2)(3Q/4)}{r^{2} }

how \frac{KQ^{2} }{r^{2} } = 0.42

Then

F = \frac{0.42*3}{8}

F = 0.1575 N

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The force exerted by the laser beam on a completely absorbing target is 3.5 \times 10^{-6} \ N.

The given parameters;

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The speed of the emitted laser light is given as;

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The force exerted by the laser beam on a completely absorbing target is calculated as follows;

P = Fv

F = \frac{P}{v} \\\\F = \frac{1050}{3\times 10^8} \\\\F = 3.5 \times 10^{-6} \ N

Thus, the force exerted by the laser beam on a completely absorbing target is 3.5 \times 10^{-6} \ N.

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