Two identical metal spheres charged with +12 | vedclass

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Question

(C) The initial electrostatic force between the spheres is given by Coulomb's law: $F_{	ext{initial}} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_1 q_2|

Vedclass · Accepted Answer

$24$

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(C) The initial electrostatic force between the spheres is given by Coulomb's law: $F_{	ext{initial}} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_1 q_2|}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{|12 	imes (-8)|}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{96}{r^2}$.When the spheres are brought into contact,the total charge is redistributed equally because the spheres are identical: $q_{	ext{new}} = \frac{q_1 + q_2}{2} = \frac{12 + (-8)}{2} = \frac{4}{2} = 2 \mu C$.After contact,the new electrostatic force is: $F_{	ext{final}} = \frac{1}{4 \pi \varepsilon_0} \frac{|q_{	ext{new}} q_{	ext{new}}|}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{2 	imes 2}{r^2} = \frac{1}{4 \pi \varepsilon_0} \frac{4}{r^2}$.The ratio of the magnitudes of the forces is: $\frac{|F_{	ext{initial}}|}{|F_{	ext{final}}|} = \frac{96/r^2}{4/r^2} = \frac{96}{4} = 24$.Thus,the ratio is $24: 1$.

Two identical metal spheres charged with $+12 \mu C$ and $-8 \mu C$ are kept at a certain distance in air. They are brought into contact and then kept at the same distance. The ratio of the magnitudes of electrostatic forces between them before and after contact is (in $: 1$)

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