Question

Two charged beads are on the plastic ring in Fig. 22-44a. Bead 2, which is not shown, is fixed in place on the ring, which has radius$\mathbf{R}\mathbf{}\mathbf{=}\mathbf{60}\mathbf{.0}\mathbf{}\mathbf{\text{\hspace{0.17em}cm}}$. Bead 1, which is not fixed in place, is initially on the x-axis at angle$\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{°}$. It is then moved to the opposite side, at angle$\mathit{\theta }\mathbf{}\mathbf{=}\mathbf{180}\mathbf{°}$, through the first and second quadrants of the x-ycoordinate system. Figure 22-44bgives the xcomponent of the net electric field produced at the origin by the two beads as a function of, and Fig. 22-44cgives the ycomponent of that net electric field. The vertical axis scales are set by ${\mathit{E}}_{\mathbf{x}\mathbf{s}}\mathbf{=}\mathbf{}\mathbf{5.0}\mathbf{}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{\text{\hspace{0.17em}N/C}}$and${\mathit{E}}_{\mathbf{y}\mathbf{s}}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{9.0}\mathbf{}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{4}}\mathbf{\text{\hspace{0.17em}N/C}}$. (a) At what angle u is bead 2 located? What are the charges of (b) bead 1 and (c) bead 2?

Short answer

1. The angle at which bead 2 is located is at $-90°$.
2. The charge of bead 1 is$2.0\times {10}^{-6}\text{C}$
3. The charge of bead 2 is$-1.6\times {10}^{-6}\text{C}$

Step-by-step solution

The given data

1. Radius of the ring,$R=60.0\text{cm}$
2. Bead 1 is initially at an angle,$\theta =0°$
3. Bead 1 is then at an angle,$\theta =180°$through first and second quadrants of the system.
4. Vertical axis scales are at $E_{xs}=5.0\times {10}^4\text{N/C}$, and $E_{ys}=-9.0\times {10}^4\text{N/C}$.

Understanding the concept of electric field 

A vector field in which any charged body will experience a force when placed in it, is known as an electric field. It points in the direction of force. The addition of the two fields gives the net electric field at a point.

The electric field is given as,

$\overrightarrow{\mathrm{E}}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2}\widehat{\mathrm{R}}$ (i)

where, $R$= The distance of field point from the charge and $q$= charge of the particle

a) Calculation of the angle at which bead 2 is located

From figure 22-44b the x component of the net electric field is zero when the value of$\theta$ is$90°$ or when bead 1 is at the point of intersection of the circle with the positive side of the y-axis. This is only possible if the bead 2 is located on the y-axis. The circular path intersects the y-axis at two points only (one along positive side of y and the other along negative side of y-axis). As the point on positive side of y-axis is already occupied by bead 1 so, bead 2 must be located at the intersection point of circular path with negative side of y-axis. Thus, the angle at which bead 2 is located is $-90°$.

b) Calculation of the charge of bead 1

Since the downward component of the net field, when bead 1 is on the +y-axis, is of the largest magnitude, then bead 1 must be a positive charge (so that its field is in the same direction as that of bead 2, in that situation). This check with the 180° value from thegraph further confirms that bead 1 is positively charged.

Thus, the charge of the bead 1 is given using equation (i) as follows:

$\begin{array}{c}{\mathrm{q}}_1=4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2\mathrm{E}\\ =\frac{(5.00\times {10}^4\text{\hspace{0.17em}N/C}){\left(0.60\text{m}\right)}^2}{\left(8.99\times {10}^9\text{N}\cdot \frac{{\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\right)}\\ =2.0\times {10}^{-6}\text{\hspace{0.17em}C}\end{array}$

Hence, the value of the charge is $2.0\times {10}^{-6}\text{\hspace{0.17em}C}$.

c) Calculation of the charge of bead 2

Similarly, the 0° value from the$E_y$graph allows us to solve for the charge of bead 2. Thus, using equation (i), the charge of the bead 2 is given as:

$\begin{array}{c}{\mathrm{q}}_2=4{\mathrm{\pi \epsilon }}_{\mathrm{o}}{\mathrm{R}}^2\mathrm{E}\\ =\frac{(-4.00\times {10}^4\text{\hspace{0.17em}N/C}){\left(0.60\text{\hspace{0.17em}m}\right)}^2}{(8.99\times {10}^9\text{N}\cdot \frac{{\text{m}}^2}{{\text{C}}^{\text{2}}})}\\ =-1.6\times {10}^{-6}\text{C}\end{array}$

Hence, the value of the charge is$-1.6\times {10}^{-6}\text{\hspace{0.17em}C}$