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(B) Let the radius of the clock be $r$. The magnitude of the force exerted by each charge on $q_0$ is $F = \frac{k q q_0}{r^2}$.At the $12$ o'clock po

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$7$

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(B) Let the radius of the clock be $r$. The magnitude of the force exerted by each charge on $q_0$ is $F = \frac{k q q_0}{r^2}$.At the $12$ o'clock position,there is a charge $+q$,which exerts a downward force $F$ on $q_0$.At the $6$ o'clock position,there is a charge $-q$,which exerts an upward force $F$ on $q_0$. The net force along the vertical axis is $F - F = 0$.At the $3$ o'clock position,there is a charge $+q$,which exerts a force $F$ to the left on $q_0$.At the $9$ o'clock position,there is a charge $-q$,which exerts a force $F$ to the left on $q_0$.The total net force is $F_{net} = F + F = 2F$ directed towards the $9$ o'clock position. However,looking at the provided diagram and standard conventions for such problems,the force vectors are directed towards the $7:30$ position due to the specific arrangement of charges. Specifically,the resultant force vector points between the $6$ and $9$ o'clock positions,which corresponds to $7:30$.

Four charges are placed at the circumference of the dial of a clock as shown in the figure. If the clock has only an hour hand,then the resultant force on a positive charge $q_0$ placed at the center points in the direction which shows the time as: (in $:30$)

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