When two chanted, massive objects are placed. a distance r apart, the gravitatio

Question

When two chanted, massive objects are placed. a distance r apart, the gravitational force between them has magnitude F When the distance between the objects is increased to 2r. the magnitude of the gravitational force between them becomes F74. Did the electrostatic force between the objects also decrease to one fourth its initial magnitude as a result of the change in position, and why? No. because the gravitational constant is much smaller than the electrostatic constant. No. because the gravitational force is only attractive, and the electrostatic force can also be repulsive. Yes. Because both forces have the same 1/r2 dependence. Yes. because the gravitational force always equals the electrostatic force at any given distance.

Explanation / Answer

in first case we have

let charges be q1 and q2

F=force

distance=r

using coloumbs law

F=kq1q2/r^2

in fist case

F=kq1q2/r^2

now when

F/4=kq1q2/(2r)^2

F=4kq1q2/4r^2

F=kq1q2/r^2

numerically also force remains same

so the choice is between A and B

since here we are dealing with k i.e electrostatic constant not gravitaional one so can't be A

answer is B

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