Three point charges +q,+2q,and +4q are placed | vedclass

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(D) Let the distance between $+q$ and $+2q$ be $r$,and the distance between $+2q$ and $+4q$ be $r$. Thus,the distance between $+q$ and $+4q$ is $2r$.T

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$1: 3$

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(D) Let the distance between $+q$ and $+2q$ be $r$,and the distance between $+2q$ and $+4q$ be $r$. Thus,the distance between $+q$ and $+4q$ is $2r$.The net electrostatic force $F_1$ on charge $+q$ is the sum of forces due to $+2q$ and $+4q$:$F_1 = \frac{k(q)(2q)}{r^2} + \frac{k(q)(4q)}{(2r)^2} = \frac{2kq^2}{r^2} + \frac{4kq^2}{4r^2} = \frac{2kq^2}{r^2} + \frac{kq^2}{r^2} = \frac{3kq^2}{r^2}$.The net electrostatic force $F_2$ on charge $+4q$ is the sum of forces due to $+2q$ and $+q$:$F_2 = \frac{k(4q)(2q)}{r^2} + \frac{k(4q)(q)}{(2r)^2} = \frac{8kq^2}{r^2} + \frac{4kq^2}{4r^2} = \frac{8kq^2}{r^2} + \frac{kq^2}{r^2} = \frac{9kq^2}{r^2}$.The ratio of the net electrostatic force on charges $+q$ and $+4q$ is:$\frac{F_1}{F_2} = \frac{3kq^2/r^2}{9kq^2/r^2} = \frac{3}{9} = 1:3$.

Three point charges $+q$,$+2q$,and $+4q$ are placed along a straight line such that the charge $+2q$ is equidistant from the other two charges. The ratio of the net electrostatic force on charges $+q$ and $+4q$ is

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