Having An Electric Time

Since the radius is constant, the component of the electric field in each direction is proportional to the charge. Thus, for some constant $k$ and for all $1 \le i \le 12$ , the electric field component towards numeral $i$ will be equal to $ki$ .

We can combine the components in opposite directions. In the direction of numeral 12, the component is $12k - 6k = 6k$ . In the direction of numeral 11, it is $11k - 5k = 6k$ . Similarly for the component in the direction of numerals 10, 9, 8, and 7. Thus we have that the electric field is the sum of six components, each of magnitude $6k$ , in the directions of numerals 7, 8, 9, 10, 11, and 12. By symmetry, the net electric field will point halfway between 9 and 10, or at $\boxed{9:30}$ .

The field of the Opposite directional charges demolish themselves . Its like -12-(-6)=-6 .So -1,-2,-3,-4,-5,-6 becomes 0 and -7,-8,-9,-10,-11,-12 becomes -6 . Next , now they are all equally charged and same direction . so find the middle number between 7 and 12 . its 9.30 . Those were all negative charges , so direction towards 9.30 . If they were +ve , it would be opposite to 9.30 ; ie 3.30 :)

Interesting question! Consider the charges $q_i$ be analogous to mass, the problem is solve for the center of 'charge' $(\bar{x},\bar{y})$ from the center of the clock $(0,0)$ as follows:

$\bar{x}= \frac{12+6}{12+6}(0) +\frac{1+5-7-11}{1+5+7+11} \cos{60^\circ} + \frac{2+4-8-10} {2+4+8+10} \cos{30^\circ} +\frac{3-9}{3+9}(1)$

$\quad = -\frac{1}{4} -\frac{\sqrt{3}}{4} -\frac{1}{2} = -\frac{1}{4}(3+\sqrt{3})$

Similarly, we find that $\bar{y}=\frac{1}{3}$ .

Let $\tan{\theta}=\frac{\bar{y}}{\bar{x}} \Rightarrow \theta = -15.7^\circ \Rightarrow$ the hour hand pointed to $\boxed {9:30}$ .

1. draw diagram
2. use vector method
3. you will get answer

check the direction of electric field ,thats it

This is a nice but easy problem from DCP.Innovative problem.

by symmetry argument we ultimately end up with two vectors along 9 direction and 12 direction, but the former is longer than the latter thus the resultant is slightly below the central position which has to be 9:30.

As the radius is constant in all cases , we can assume the the net charge is concentrated at one place i.e. center of charge ( just like center of mass). And at that point the net electric field is acting.