Four charges are placed at circumference of dial of clock as shown in figure. If the clock has only hour hand, then the resultant force on a positive charge q0 placed at centre, points in the direction which shows time as

Dear Student,

Ans (2) 7:30

The magnitude of force on q_o due to each of the 4 charges will be equal as they all have same magnitude and all are at same distance from q_{o  .} So let this force be ${F_q}_{-q_0} (magnitude of force due to each charge on q_o)$
Now lets see directions of all  4 forces on q_o(positive)
due to cahrge at 

a) 9 and 3 -- towards 9 (attractive and repulsive respectively)

b) 12 and 6 -- towards 6 (repulsive and attractive respectively)

so net force -  towards 9 is $2F_{q-q_0}$ and towards 6 is also  $2F_{q-q_0}$

now we are left with finding resultant of these to forces only

Now as both the forces are equal in magnitude and perpendicular to each other so there resultant will be exactly  midway between 6 and 9  i.e. 7:30  
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