Q17P - Fundamentals Of Physics

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-y coordinate system. Figure 22-44b gives the x component of the net electric field produced at the origin by the two beads as a function of, and Fig. 22-44c gives the y component 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?

Step-by-Step Solution

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 1: 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 $E_{ys} =-9.0 \times {10}^4\text{ N/C}$.

Step 2: 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

Step 3: 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 $-90°$.

Step 4: 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 the  graph 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 $2.0 \times {10}^{-6}\text{\hspace{0.17em}C}$.

Step 5: 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$-1.6 \times {10}^{-6}\text{\hspace{0.17em}C}$