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(A) Let the side length of the square be $r$. The charges at vertices $1, 2,$ and $3$ are identical,so $q_1 = q_2 = q_3 = q$.The distance between char

Vedclass · Accepted Answer

$\frac{1}{2}$

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(A) Let the side length of the square be $r$. The charges at vertices $1, 2,$ and $3$ are identical,so $q_1 = q_2 = q_3 = q$.The distance between charges $q_1$ and $q_2$ is the side of the square,$r_{12} = r$.According to Coulomb's Law,the force $F_{12}$ is:$F_{12} = \frac{k q_1 q_2}{r_{12}^2} = \frac{k q^2}{r^2} \quad \dots (1)$The distance between charges $q_1$ and $q_3$ is the diagonal of the square,$r_{13} = \sqrt{r^2 + r^2} = r\sqrt{2}$.According to Coulomb's Law,the force $F_{13}$ is:$F_{13} = \frac{k q_1 q_3}{r_{13}^2} = \frac{k q^2}{(r\sqrt{2})^2} = \frac{k q^2}{2r^2} \quad \dots (2)$Taking the ratio of equation $(2)$ to equation $(1)$:$\frac{F_{13}}{F_{12}} = \frac{\frac{k q^2}{2r^2}}{\frac{k q^2}{r^2}} = \frac{1}{2}$Thus,the correct option is $A$.

Three identical charges are placed on three vertices of a square. If the force acting between $q_1$ and $q_2$ is $F_{12}$ and between $q_1$ and $q_3$ is $F_{13}$,then $\frac{F_{13}}{F_{12}} = $ . . . . . . .

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