Four charges are placed at the circumference | vedclass

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(B) Let the radius of the clock be $R$. The charges are placed at the positions of $12, 3, 6,$ and $9$ o'clock.At $12$ o'clock, charge is $+q$. At $6$

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$7$

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(B) Let the radius of the clock be $R$. The charges are placed at the positions of $12, 3, 6,$ and $9$ o'clock.At $12$ o'clock, charge is $+q$. At $6$ o'clock, charge is $-q$. At $3$ o'clock, charge is $+q$. At $9$ o'clock, charge is $-q$.The force on $q_0$ at the center due to $+q$ at $12$ is $F_1 = \frac{kq_0q}{R^2}$ (directed towards $6$).The force on $q_0$ at the center due to $-q$ at $6$ is $F_2 = \frac{kq_0q}{R^2}$ (directed towards $6$).Total force along the vertical axis is $F_y = F_1 + F_2 = \frac{2kq_0q}{R^2}$ (downwards towards $6$).The force on $q_0$ at the center due to $+q$ at $3$ is $F_3 = \frac{kq_0q}{R^2}$ (directed towards $9$).The force on $q_0$ at the center due to $-q$ at $9$ is $F_4 = \frac{kq_0q}{R^2}$ (directed towards $9$).Total force along the horizontal axis is $F_x = F_3 + F_4 = \frac{2kq_0q}{R^2}$ (leftwards towards $9$).The resultant force $F_{net}$ is directed at an angle of $45^{\circ}$ between the $6$ and $9$ o'clock positions.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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