Charges q, 2q, 3q, and 4q are placed at the c | vedclass

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(A) Let the corners of the square be $A, B, C,$ and $D$ in clockwise order starting from top-left,with charges $4q, 3q, 2q,$ and $q$ respectively. The

Vedclass · Accepted Answer

$2\sqrt{2} \frac{kq q_0}{r^2}$

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(A) Let the corners of the square be $A, B, C,$ and $D$ in clockwise order starting from top-left,with charges $4q, 3q, 2q,$ and $q$ respectively. The distance from each corner to the center $O$ is $r$.The force exerted by charge $q$ at $D$ on $q_0$ at $O$ is $F_D = \frac{kq q_0}{r^2}$ (directed away from $D$).The force exerted by charge $3q$ at $B$ on $q_0$ at $O$ is $F_B = \frac{k(3q) q_0}{r^2}$ (directed away from $B$).The net force along the diagonal $BD$ is $F_{BD} = F_B - F_D = \frac{3kq q_0}{r^2} - \frac{kq q_0}{r^2} = \frac{2kq q_0}{r^2}$ (directed towards $D$).Similarly,the force exerted by charge $2q$ at $C$ on $q_0$ at $O$ is $F_C = \frac{k(2q) q_0}{r^2}$ (directed away from $C$).The force exerted by charge $4q$ at $A$ on $q_0$ at $O$ is $F_A = \frac{k(4q) q_0}{r^2}$ (directed away from $A$).The net force along the diagonal $AC$ is $F_{AC} = F_A - F_C = \frac{4kq q_0}{r^2} - \frac{2kq q_0}{r^2} = \frac{2kq q_0}{r^2}$ (directed towards $C$).Since the diagonals of a square are perpendicular,the net force $F_{net}$ is the vector sum of $F_{BD}$ and $F_{AC}$,which are equal in magnitude $(F = \frac{2kq q_0}{r^2})$ and perpendicular to each other.$F_{net} = \sqrt{F^2 + F^2} = F\sqrt{2} = \frac{2kq q_0}{r^2} 	imes \sqrt{2} = 2\sqrt{2} \frac{kq q_0}{r^2}$.

Charges $q, 2q, 3q,$ and $4q$ are placed at the corners of a square as shown in the figure. $A$ charge $q_0$ is placed at its center. Find the net force on the charge $q_0$.

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