91Ó°ÊÓ

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

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,

$\stackrel{→}{\mathrm{E}}=\frac{\mathrm{q}}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}{\mathrm{R}}^2}\widehat{\mathrm{R}}$ (i)

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

From figure 22-44b the x component of the net electric field is zero when the value of$\mathrm{θ}$ 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 $-90°$.

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:

$\begin{array}{c}{\mathrm{q}}_1=4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}{\mathrm{R}}^2\mathrm{E}\\ =\frac{(5.00×{10}^4\text{ N/°ä}){\left(0.60\text{m}\right)}^2}{\left(8.99×{10}^9\text{ }\text{N}â‹\dots \frac{{\text{m}}^{\text{2}}}{{\text{C}}^{\text{2}}}\right)}\\ =2.0×{10}^{−6}\text{ C}\end{array}$

Hence, the value of the charge is $2.0×{10}^{−6}\text{ C}$.

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:

$\begin{array}{c}{\mathrm{q}}_2=4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}{\mathrm{R}}^2\mathrm{E}\\ =\frac{(−4.00×{10}^4\text{ N/°ä}){\left(0.60\text{″¾}\right)}^2}{(8.99×{10}^9\text{N}â‹\dots \frac{{\text{m}}^2}{{\text{C}}^{\text{2}}})}\\ =−1.6×{10}^{−6}\text{C}\end{array}$

Hence, the value of the charge is$−1.6×{10}^{−6}\text{ C}$