Four charges,each of magnitude +Q,are placed | vedclass

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(C) Let the charges be at corners $A, B, C,$ and $D$ of a square with side $a$. We calculate the net force on the charge at corner $B$.$1$. The force

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the charges be at corners $A, B, C,$ and $D$ of a square with side $a$. We calculate the net force on the charge at corner $B$.$1$. The force due to charge at $A$ is $F_A = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{a^2}$ (along $AB$).$2$. The force due to charge at $C$ is $F_C = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{a^2}$ (along $CB$).$3$. The resultant of $F_A$ and $F_C$ is $F_{AC} = \sqrt{F_A^2 + F_C^2} = \sqrt{2} \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{a^2}$.$4$. The force due to charge at $D$ is $F_D = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{(a\sqrt{2})^2} = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{2a^2}$ (along $DB$).$5$. The net force $F_{net} = F_{AC} + F_D = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) = \frac{1}{4\pi \varepsilon_0} \frac{Q^2}{a^2} \left( \frac{2\sqrt{2} + 1}{2} \right)$.

Four charges,each of magnitude $+Q$,are placed at the corners of a square of side $a$. What is the net force on one of the charges?

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