91影视

(a) To determine which ball has the positive charge, we will follow the direction of the electric field E is directed to the left, which means that the lines of the field are directed to the left. Lines of the field always go from the positively charged side to the negatively charged, so we can conclude that the positive charge that causes the field E is on the right side and the negative charge is on the left side.

According to Coulomb's law, the force that acts between two charged bodies is as such that two charges of the same sign go from each other and two charges of the opposite sign go towards each other. Since the positive charge of the electric field E is on the right side, the positively charged ball is on the left side

(b) To determine the magnitude of the field, we will use Newton's second law. We will consider only the forces that act upon the negatively charged ball.

• ${\mathrm{F}}_{\mathrm{g}}=\mathrm{mg}$gravitational force, which is directed downwards,

• ${\mathrm{F}}_{\mathrm{e}}=\frac{{\mathrm{kq}}^2}{{\mathrm{r}}^2}$Coulomb's force between the balls, which is directed to the left,

• $F_E=q_E$Force of the electric field, which is directed to the right and

• T tension of the string, which is directed along the string

The sum of these forces is equal to zero since the ball is in the equilibrium state

In the horizontal direction, using Newton's second law, we can write

$\frac{{\mathrm{kq}}^2}{{\mathrm{r}}^2}+\mathrm{T}\sin \frac{\mathrm{胃}}{2}-\mathrm{qE}=0$

in the vertical direction, using Newton's second law, we can write

$\mathrm{T}\cos \frac{\mathrm{胃}}{2}-\mathrm{mg}=0$

Combining these two equations, to get rid of T, we get

$$\begin{gathered} \mathrm{T}\sin \frac{\mathrm{胃}}{2}=\mathrm{qE}-\frac{{\mathrm{kq}}_2}{{\mathrm{r}}^2} \\ \tan \frac{\mathrm{胃}}{2}=\frac{\mathrm{qE}-{\displaystyle \frac{{\mathrm{kq}}^2}{{\mathrm{r}}^2}}}{\mathrm{mg}} \end{gathered}$$

From this equation, we can now calculate E

$$\begin{gathered} \tan \frac{\mathrm{胃}}{2}=\frac{\mathrm{qE}-{\displaystyle \frac{{\mathrm{kq}}^2}{{\mathrm{r}}^2}}}{\mathrm{mg}} \\ \tan \frac{\mathrm{胃}}{2}.\mathrm{mg}=\mathrm{qE}-\frac{{\mathrm{kq}}^2}{{\mathrm{r}}^2} \\ \mathrm{E}=\tan \frac{\mathrm{胃}}{2}.\frac{\mathrm{mg}}{\mathrm{q}}+\mathrm{q}+\frac{{\mathrm{kq}}^2}{{\left(2\mathrm{Isin}{\displaystyle \frac{\mathrm{胃}}{2}}\right)}^2} \end{gathered}$$

Putting the given values into the equation for E, we get:

$$\begin{gathered} \mathrm{E}=\tan \frac{\mathrm{胃}}{2}.\frac{\mathrm{mg}}{\mathrm{q}}+\mathrm{q}+\frac{{\mathrm{kq}}^2}{{\left(\mathrm{Isin}{\displaystyle \frac{\mathrm{胃}}{2}}\right)}^2} \\ =\tan \left({29}^{掳}\right).\frac{6.8.{10}^{-6}\mathrm{kg}.9.81\mathrm{m}/{\mathrm{s}}^2}{72.{10}^{-9}\mathrm{C}}+\frac{9.{10}^9{\mathrm{Nm}}^2{\mathrm{C}}^2.72.{10}^{-9}\mathrm{C}}{{\left(2.0.530\mathrm{m}.\sin \left({29}^{掳}\right)\right)}^2} \end{gathered}$$

Result:

(a) Left

(b) E= 2963.8 N/C