A clock face has charges –q, –2q, –3q, ..... –12q fixed at the position of the corresponding numerals on the dial. The clock hands do not disturb the net field due to point charges. At what time does the hour hand point in the direction of the electric field at the centre of the dial.

The net electric field due to the oppositely faced charges would be in the direction of electric fields produced by −7q,−8q,−9q,−10q,−11q and −12q and the magnitude of these electric fields would be $6E_{1 }$each, where $E_1 is the electric field produced by the charge q at 1.$
 The resultant of these fields has two mutually perpendicular components, one is directed towards -9q charge and has magnitude $(2+\sqrt{3}) 6E_1.$ The other is directed towards -12q charge and has magnitude $6E_{1 }$.
Now, finally the resultant of $(2+\sqrt{3})6E_1 and 6E_{1}makes an angle \theta with the direction of (2+\sqrt{3})6E_1$,where $\tan \theta =\frac{6E_1}{(2+\sqrt{3})6E_1}\Rightarrow \theta = {15}^0$ and would point in the direction of 9:30.