As shown in the figure,charges +q each are pl | vedclass

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(A) Let the vertices of the square be $A, B, C,$ and $D$ in order. The charges at $A, B, C$ exert forces on the charge at $D$.$1$. Force due to charge

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the vertices of the square be $A, B, C,$ and $D$ in order. The charges at $A, B, C$ exert forces on the charge at $D$.$1$. Force due to charge at $A$ $(F_{DA})$: This force acts along $DA$ (upwards) with magnitude $F_{DA} = \frac{k q^2}{a^2}$.$2$. Force due to charge at $C$ $(F_{DC})$: This force acts along $DC$ (leftwards) with magnitude $F_{DC} = \frac{k q^2}{a^2}$.$3$. Force due to charge at $B$ $(F_{DB})$: The distance between $B$ and $D$ is the diagonal of the square,which is $\sqrt{2}a$. The force acts along the diagonal $BD$ (outwards) with magnitude $F_{DB} = \frac{k q^2}{(\sqrt{2}a)^2} = \frac{k q^2}{2a^2}$.The resultant force of $F_{DA}$ and $F_{DC}$ is $F_{AC} = \sqrt{F_{DA}^2 + F_{DC}^2} = \sqrt{\left(\frac{k q^2}{a^2}\right)^2 + \left(\frac{k q^2}{a^2}\right)^2} = \frac{\sqrt{2} k q^2}{a^2}$. This resultant force acts along the diagonal $BD$.Since $F_{AC}$ and $F_{DB}$ are in the same direction,the net force is $F_{net} = F_{AC} + F_{DB} = \frac{\sqrt{2} k q^2}{a^2} + \frac{k q^2}{2a^2} = \left(\sqrt{2} + \frac{1}{2}\right) \frac{k q^2}{a^2}$.

As shown in the figure,charges $+q$ each are placed at the four vertices of a square of side length $a$. The Coulomb force acting on the charge placed at vertex $D$ is . . . . . . .

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