Two point charges of 40 mu C and -20 mu C a | vedclass

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(A) The electrostatic force between two point charges is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2}$.Since the distance $r$ remains the same

Vedclass · Accepted Answer

$8:1$

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(A) The electrostatic force between two point charges is given by Coulomb's Law: $F = k \frac{|q_1 q_2|}{r^2}$.Since the distance $r$ remains the same,the force is proportional to the product of the charges: $F \propto |q_1 q_2|$.In the first case,the product of the charges is $|q_1 q_2| = |40 	imes (-20)| = 800 \ \mu C^2$.When the two charges are brought into contact,the total charge is shared equally between them (assuming identical conductors): $q' = \frac{q_1 + q_2}{2} = \frac{40 - 20}{2} = 10 \ \mu C$.In the second case,the product of the charges is $|q'_1 q'_2| = |10 	imes 10| = 100 \ \mu C^2$.The ratio of the forces is $\frac{F_1}{F_2} = \frac{|q_1 q_2|}{|q'_1 q'_2|} = \frac{800}{100} = \frac{8}{1}$.

Two point charges of $40 \ \mu C$ and $-20 \ \mu C$ are placed at a certain distance. If they are brought into contact and then placed at the same distance,what is the ratio of the electrostatic force in the two cases?

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