Four charges Q_1, Q_2, Q_3, and Q_4 of the sa | vedclass

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Question

(A) Let the force exerted by charge $Q_i$ on $q$ be $\vec{F}_i$. The $x$-component of the force from a pair of charges at $\pm x_0$ is proportional to

Vedclass · Accepted Answer

$P-3, Q-1, R-4, S-2$

Vedclass · Answer

(A) Let the force exerted by charge $Q_i$ on $q$ be $\vec{F}_i$. The $x$-component of the force from a pair of charges at $\pm x_0$ is proportional to $(Q_{left} - Q_{right})$. The $y$-component is proportional to $(Q_{left} + Q_{right})$.$(P)$ All positive: $Q_1=Q_2=Q_3=Q_4 = +Q$. The $x$-components cancel out due to symmetry. The $y$-components add up, resulting in a net force along $+y$ (Option $3$).$(Q)$ $Q_1, Q_2$ positive, $Q_3, Q_4$ negative: The $y$-components cancel out. The $x$-components add up in the $+x$ direction because the positive charges are on the left and negative on the right, pushing $q$ to the right (Option $1$).$(R)$ $Q_1, Q_4$ positive, $Q_2, Q_3$ negative: The $x$-components cancel out. The $y$-components are negative because the charges closer to the origin $(Q_2, Q_3)$ are negative and stronger in their $y$-effect than the farther positive charges $(Q_1, Q_4)$, resulting in a net force along $-y$ (Option $4$).$(S)$ $Q_1, Q_3$ positive, $Q_2, Q_4$ negative: The $y$-components cancel out. The $x$-components result in a net force along $-x$ because the net effect of the charges on the left is negative and on the right is positive, pulling $q$ to the left (Option $2$).Thus, the correct matching is $P-3, Q-1, R-4, S-2$.

List-$I$	List-$II$
$P. Q_1, Q_2, Q_3, Q_4$ all positive	$1. +x$
$Q. Q_1, Q_2$ positive; $Q_3, Q_4$ negative	$2. -x$
$R. Q_1, Q_4$ positive; $Q_2, Q_3$ negative	$3. +y$
$S. Q_1, Q_3$ positive; $Q_2, Q_4$ negative	$4. -y$

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