Physics, 25.07.2020 01:01 roofroofmom

When separated by distance d, identically charged point-like objects A and B exert a force of magnitude F on each other. If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, what will be the new force that they exert on each other in terms of force F

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Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. If the charges are reduced to one-fourth of their original values and the distance is halved, the new force will be one-fourth of the original force.

Two identically charged point-like objects A and B exert a force of magnitude F on each other when separated by distance d. This can be explained through Coulomb's law.

What is Coulomb's law?

Coulomb's law is a law stating that like charges repel and opposite charges attract, with a force proportional to the product of the charges and inversely proportional to the square of the distance between them.

$F = k \frac{q_Aq_B}{d^{2} } = k \frac{q^{2} }{d^{2} }$

where,

and

are the charges of A and B (and equal to q).k is the Coulomb's constant.

If you reduce the charge of A to one-fourth its original value, and the charge of B to one-fourth, and reduce the distance between the objects by half, the new force will be:

$F_2 = k \frac{(0.25q_A)(0.25q_B)}{(0.5d)^{2} } = 0.25k\frac{q^{2} }{d^{2} } = 0.25 F$

Learn more about Coulomb's law here: link

Answer from: junkmailemail42

F ’= F 0.25

Explanation:

This problem refers to the electric force, which is described by Coulomb's law

F = k q₁ q₂ / r²

where k is the Coulomb constant, q the charges and r the separation between them.

The initial conditions are

F = k q_A q_B / d²

they indicate that the loads are reduced to ¼ q and the distance is reduced to ½ d

F ’= k (q / 4 q / 4) / (0.5 d)²

F ’= k q / 16 / 0.25 d²

F ’= k q² / d² 0.0625 / 0.25

F ’= F 0.25

Answer from: Quest

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Answer from: Quest

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