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If three equal charges are placed on three corners of square and force between $q_1$ and $q_2$ is $F_2$, and the force between $q_1$ and $q_2$ is $F_3$, find $F_3/F_2$

$\begin{array}{1 1} q_3 /q_2 \\ q_3.q_2 \\ 2 \\ 1/2 \end{array}$

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  • Coulomb's law states that the magnitude of the electrostatic force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distance between them
  • Force $F = \large\frac{ k_e q_1 q_2 }{ r^2}$, where $k_e$ is Coulumb's constant, $q_1$ and $q_2$ signed magnitudes of the charges, the scalar $r$ is the distance between the charges,
Let the side of a square be $a$, where $q_1 = q_2 = q_3 = q_4$
We know that $F = \large\frac{ k_e q_1 q_2 }{ r^2}$ and since $q_1 = q_2 = q_3 = q_4$, $F_3 = \large\frac{ k_e q q }{ (a \sqrt 2)^2} = \frac{ k_e q^2 }{ 2a^2}$
Similarly, $F_2 = \large\frac{ k_e q q }{ a^2} = \frac{ k_e q^2 }{ a^2}$
Therefore $\large \frac{F_3}{F_2} =$$ \large\frac{ \frac{ k_e q^2 }{ 2a^2}} {\frac{k_e q^2}{a^2}} $
$\Rightarrow \large \frac{F_3}{F_2} =\large \frac{1}{2}$
answered Dec 20, 2013 by balaji.thirumalai

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