91Ó°ÊÓ

1. Radius of the ring, $R=50.0\text{cm or}0.5\text{m}$
2. Charge of bead 1 which is at the left side, $q_1=+2\mathrm{μ}\text{C or}+\text{2}×{\text{10}}^{-6}\text{C}$
3. Charge of bead 2 that can be moved along the ring, $q_2=+6\mathrm{μ}\text{C or}+6×{\text{10}}^{-6}\text{C}$
4. The value of electric field, $E=2.00×{10}^5\text{N/C}$

The field due to a positively charged body at a point is outward that is moving away from the point. This direction is given by the concept of force considering that a positive charge body is located at the centre. Thus, the field directions will be opposite due to positive charges located at the circumference of the ring and the net field will have two components due to the angle made by bead 2.

Here, the electric field due to the present charges in the system on the central particle is likely to have positive and negative angles of equal magnitude as the central particle on which the field is affecting, and one of the beads that are the charged particle is located on the same x-axis, while the other charged bead has a component of its electric field along the x-axis in the opposite direction of the field due to the former charge.

Formulae:

The electric field at a point due to a point charge, $\stackrel{→}{E}=\frac{q}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2}\hat{R}$ (i)

where, $k=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0}=9×{10}^9{\text{N.m}}^2/{\text{C}}^2$is the constant value, $\hat{R}$ = The distance of field point from the charge, q = charge of the particle, is the unit vector of the radial distance.

The magnitude of a vector, ${\left|A\right|}^2=A_x^2+A_y^2$ (ii)

where, $\stackrel{→}{A},$is the horizontal component of vector $\stackrel{→}{A},A_y$is the vertical component of vector .

The net field components along the x and y axes using equation (i) and the given figure can be given as:

$$\begin{gathered} E_{net,x}=\frac{q_1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2}-\frac{q_{2}cos\mathrm{θ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2}, \\ {E}_{net,y}=-\frac{q_{2}sin\mathrm{θ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2} \end{gathered}$$

Now, the magnitude of the net electric field can be given using equation (ii) and the given data as follows:

$$\begin{gathered} E^2={\left(\frac{q_1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2}-\frac{q_{2}cos\mathrm{θ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2}\right)}^2+{\left(-\frac{q_{2}sin\mathrm{θ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2}\right)}^2 \\ =\frac{q_1^2}{{\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2\right)}^2}+\frac{q_2^2{\cos }^2\mathrm{θ}}{{\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2\right)}^2}-\frac{2q_1{q}_2}{{\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2\right)}^2}+\frac{q_2^2{\sin }^2\mathrm{θ}}{{\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2\right)}^2} \\ =\frac{q_1^2+q_2^2-2q_1{q}_{2}cos\mathrm{θ}}{{\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_o{R}^2\right)}^2} \\ (2.00×{10}^5\text{N/C}{)}^2=\frac{{(2×{10}^{-6}\text{C})}^2+{(6×{10}^{-6}\text{C})}^2-2(2×{10}^{-6}\text{C})(6×{10}^{-6}\text{C})\cos \mathrm{θ}}{{\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_o{(0.5\text{m})}^2\right)}^2} \\ \mathrm{θ}=co{s}^{-1}\left\{\frac{{\left(4\mathrm{Ï€}{\mathrm{Î\mu }}_o{(0.5\text{m})}^2\right)}^2×{(2.00×{10}^5\text{N/C})}^2}{{(2×{10}^{-6}\text{C})}^2+{(6×{10}^{-6}\text{C})}^2-2(2×{10}^{-6}\text{C})(6×{10}^{-6}\text{C})}\right\} \\ =67.8^0\text{or}-67.8^0 \end{gathered}$$

Hence, the positive value of the angle is 67.8⁰

From the above calculations of part (a), it can be said that the negative value of the angle is$-67.8^0$

Hence, the negative angle at which the bead 2 is positioned is $-67.8^0$.