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Here attractive force will workSo if we assume a test charge placed at centre then for numbers of right side means in opposite of -12q we have -6q so half of the force will be cancelled and the net due to both will be in direction of -12q charge.Similarly this will be followed by other number too.So the the resultant of all net forces will be somewhere b/w no. 9 and 10.So at 9:30 the hour hand will direct towards the direction of net Electric field.I hope this helps. Thanks a lot
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