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(B) Let the center of the clock be the origin $(0,0)$.At $12$ o'clock position,charge is $+q$.At $3$ o'clock position,charge is $+q$.At $6$ o'clock po

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$7$

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(B) Let the center of the clock be the origin $(0,0)$.At $12$ o'clock position,charge is $+q$.At $3$ o'clock position,charge is $+q$.At $6$ o'clock position,charge is $-q$.At $9$ o'clock position,charge is $-q$.Let the force exerted by a charge $q$ on $q_0$ at the center be $F = \frac{kq q_0}{r^2}$.Force due to $+q$ at $12$ o'clock is $F$ downwards (towards $6$).Force due to $-q$ at $6$ o'clock is $F$ downwards (towards $6$).Total vertical force $F_y = F + F = 2F$ (downwards).Force due to $+q$ at $3$ o'clock is $F$ towards left (towards $9$).Force due to $-q$ at $9$ o'clock is $F$ towards left (towards $9$).Total horizontal force $F_x = F + F = 2F$ (towards left).The resultant force $F_{net} = \sqrt{F_x^2 + F_y^2} = \sqrt{(2F)^2 + (2F)^2} = 2\sqrt{2}F$.The direction of the resultant force makes an angle of $45^{\circ}$ with the vertical and horizontal axes,pointing towards the third quadrant (between $6$ and $9$ o'clock).This direction corresponds to the time $7:30$ on the clock.

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

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