A clock face has negative point charges,$\mathbf{-}\mathit{q}$,$\mathbf{-}\mathbf{2}\mathit{q}$, $\mathbf{-}\mathbf{3}\mathit{q}$. . . ,$\mathbf{-}\mathbf{12}\mathit{q}$fixed at the positions of the corresponding numerals. The clock hands do not perturb the net field due to the point charges. At what time does the hour hand point in the same direction as the electric field vector at the center of the dial? (Hint:Use symmetry.)

A clock face has negative point charges$-q,-2q,-3q,\mathrm{...},-12q$ fixed at the positions of the corresponding numerals.

Using the concept of the electric field, we can get the required value of time on the condition that the hour hand is in the direction of the net electric field at the centre of the clock.

Formula:

The electric field is, $\overrightarrow{E}=\frac{q}{4\pi {\epsilon }_o{r}^2}\widehat{r}$ (i)

Where, r = the distance of field point from the charge

q = charge of the particle

We consider pairs of diametrically opposed charges. The net field due to just the charges in the one o’clock$(–q)$and seven o’clock$(–7q)$positions is clearly equivalent to that of a single$(–6q)$charge sitting at the seven o’clock position. Similarly, the net field due to just the charges in the six o’clock$(–6q)$and twelve o’clock$(–12q)$positions is the same as that due to a single$(–6q)$charge sitting at the twelve o’clock position. Continuing with this line of reasoning, we see that there are six equal-magnitude electric field vectors pointing at the seven o’clock, eight o’clock, till the twelve o’clock positions.

Thus, the resultant field of all of these points, by symmetry, is directed toward the position midway between the seven and the twelve o’clock. Therefore,$E_{resultant}$ points toward the nine-thirty position.