Equal charges q are placed at the four corner | vedclass

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(C) Let $k = \frac{1}{4\pi \varepsilon_0}$. The charge at $B$ experiences forces from charges at $A, C,$ and $D$.$1$. Force due to charge at $A$: $F_A

Vedclass · Accepted Answer

$\left( \frac{1 + 2\sqrt{2}}{2} \right) \frac{q^2}{4\pi \varepsilon_0 a^2}$

Vedclass · Answer

(C) Let $k = \frac{1}{4\pi \varepsilon_0}$. The charge at $B$ experiences forces from charges at $A, C,$ and $D$.$1$. Force due to charge at $A$: $F_A = \frac{kq^2}{a^2}$ (directed along $AB$ extended).$2$. Force due to charge at $C$: $F_C = \frac{kq^2}{a^2}$ (directed along $CB$ extended).$3$. Force due to charge at $D$: $F_D = \frac{kq^2}{(a\sqrt{2})^2} = \frac{kq^2}{2a^2}$ (directed along $DB$ extended).Since $F_A$ and $F_C$ are perpendicular,their resultant $F_{AC} = \sqrt{F_A^2 + F_C^2} = \sqrt{\left(\frac{kq^2}{a^2}\right)^2 + \left(\frac{kq^2}{a^2}\right)^2} = \sqrt{2} \frac{kq^2}{a^2}$.This resultant $F_{AC}$ acts in the same direction as $F_D$ (along the diagonal $DB$).Therefore,the net force $F_{net} = F_{AC} + F_D = \sqrt{2} \frac{kq^2}{a^2} + \frac{kq^2}{2a^2} = \frac{kq^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) = \frac{kq^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right)$.Substituting $k = \frac{1}{4\pi \varepsilon_0}$,we get $F_{net} = \left( \frac{1 + 2\sqrt{2}}{2} \right) \frac{q^2}{4\pi \varepsilon_0 a^2}$.

Equal charges $q$ are placed at the four corners $A, B, C, D$ of a square of side length $a$. The magnitude of the net electrostatic force on the charge at $B$ is:

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