Kaplan PS Section Test #30 - Electrostatics

[moto_za]

Would appreciate any help understanding how to go about this question:

Two identical conducting spheres on insulating supports carry charges of magnitude +Q and +2Q respectively. When they are separated by a distance of d the electrostatic repulsive force is F. The spheres are made to touch and then restored to their original separation d. If there is no loss of charge, what is the new force of repulsion in terms of F?

Answer: 9F/8

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[dudewheresmymd]

Would appreciate any help understanding how to go about this question:

Two identical conducting spheres on insulating supports carry charges of magnitude +Q and +2Q respectively. When they are separated by a distance of d the electrostatic repulsive force is F. The spheres are made to touch and then restored to their original separation d. If there is no loss of charge, what is the new force of repulsion in terms of F?

Answer: 9F/8

Original force is F=(Q)(2Q)/d^2 = 2Q^2/d^2

New Force F' = (1.5Q)(1.5Q)/d^2 = 2.25Q^2/d^2

Ratio F'/F = 2.25/2=1.125=9/8F


The concept behind this is that the total charge is (Q)+(2Q) before = 3Q when they touch the charges will be evenly distributed among the two spheres (3Q/2)= 1.5Q now on each. Compare the ratio of F' to F to get 9/8F.

 

[moto_za]

Original force is F=(Q)(2Q)/d^2 = 2Q^2/d^2

New Force F' = (1.5Q)(1.5Q)/d^2 = 2.25Q^2/d^2

Ratio F'/F = 2.25/2=1.125=9/8F


The concept behind this is that the total charge is (Q)+(2Q) before = 3Q when they touch the charges will be evenly distributed among the two spheres (3Q/2)= 1.5Q now on each. Compare the ratio of F' to F to get 9/8F.

Thanks so much! You explained it much better than Kaplan

 

[dudewheresmymd]

No problem...sometimes when I'm in kaplan class I feel like i'm actually decreasing my mcat score...and that the only thing that is helping me is their online materials...definitely not worth the 2000.