Two small charged objects repel each other with a force Fwhen separated by a distance d.If the charge on each object is reduced to one-fourth of itsoriginal value and the distance between them is...

Two small charged objects repel each other with a force Fwhen separated by a distance d.

If the charge on each object is reduced to one-fourth of itsoriginal value and the distance between them is reduced to d/2 the force become F/8

explain

Asked on by Adian

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jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

This question is related to Coulomb's Law.

Coulomb's Law

"the magnitude of the Electrostatics force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distances between them"


If we assume the magnitude of two charges are q_1 and q_2
and they are at a distance of r the repel force is F then;

F = K(q_1*q_2)/r^2 where K is a constant.

 

F represent the force of repel between the charges. When the power or magnitude of the charge decreases The Force of repel of the charges will decrease since Electrostatics force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges.

Also the charges will come closer because now the force is reduced and repelling power of the charges are low.

Sources:

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