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7nadin3 [17]
3 years ago
12

Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric

force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:
Physics
1 answer:
Goshia [24]3 years ago
3 0

Complete Question:

Metal sphere A has a charge of − Q . −Q. An identical metal sphere B has a charge of + 2 Q . +2Q. The magnitude of the electric force on sphere B due to sphere A is F . F. The magnitude of the electric force on sphere A due to sphere B must be:

A. 2F

B. F/4

C. F/2

D. F

E. 4F

Answer:

D.

Explanation:

If both spheres can be treated as point charges, they must obey the Coulomb's law, that can be written as follows (in magnitude):

F =\frac{kQ*2Q}{r^{2} }

As it can be seen, this force is proportional to the product of the charges, so it must be the same for both charges.

As this force obeys also the Newton's 3rd Law, we conclude that the magnitude of the electric force on sphere A due to sphere B, must be equal to the the magnitude of the force on the sphere B due to the sphere A, i.e., just F.

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Optical tweezers use light from a laser to move single atoms and molecules around. Suppose the intensity of light from the tweez
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Explanation:

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2 years ago
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2 years ago
A particle leaves the origin with a speed of 3 106 m/s at 38 degrees to the positive x axis. It moves in a uniform electric fiel
Salsk061 [2.6K]

Answer:

If the particle is an electron E_y = 3.311 * 10^3 N/C

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Explanation:

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\theta = 38^0 to +ve x-axis

The particle crosses the x-axis at , x = 1.5 cm = 0.015 m

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The electric field intensity along the positive y axis E_y, can be given by the formula:

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If the particle is an electron:

E_y = \frac{2 m_e u^2 sin \theta cos \theta}{qx} \\

E_y = \frac{2 * 9.1 * 10^{-31} * (3*10^6)^2 *(sin38)( cos38)}{1.6*10^{-19} * 0.015} \\

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8 0
3 years ago
The center-seeking change in velocity of an object moving in a circle is:
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The center-seeking change in velocity of an object moving in a circle is the centripetal acceleration.

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