Question

Two charged beads are on the plastic ring in figure (a). Bead $2$ , which is not shown, is fixed in place on the ring and has a radius $R = 40.0 cm .$ Bead $1,$ which is not fixed in place is initially on the $x$ -axis at angle $θ = 0 ° .$ It is then, moved to the opposite side, at angle $θ = 180 °,$ through the first and second quadrants of the $x y$ coordinate system. Figure (b) gives the $x$ -component of the net electric field produced at the origin by the two beads as a function of $θ$ , and figure (c) gives the $y$ -component of that net electric field. The vertical axis scales are set by $E_{n} = 5.0 \times 10^{4} N C^{- 1}$ and $E_{y s} = - 9.0 \times 10^{4} N C^{- 1} . (a)$ At what angle is bead $2$ located? What are the charges of $(b)$ bead $1$ and bead $2 ?$

Accepted Answer

We make the assumption that bead $2$ is in the lower half of the circle, partly because it would be awkward for bead $R = 40.0 cm .$ to "slide through" bead $2$ if it were in the path of bead $1$ (which is the upper half of the circle) and partly to eliminate a second solution to the problem (which would have opposite angle and charge for bead $2$ ). We note that the net $y$ -component of the electric field evaluated at the origin is negative (points down) for all positions of bead $1$ , which implies (with our assumption in the previous sentence) that bead $2$ is a negative charge.

$(a)$ When bead $1$ is on the $+ y$ axis, there is no $x$ -component of the net electric field, which implies that bead $2$ in on the $x y$ axis, so its angle is $x$ .

$(b)$ Since the downward component of the net field, when bead $1$ is on the $+ y$ axis, is of 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). Comparing the values of $E_{y}$ at $0 °$ and at $E_{y s} = - 9.0 \times 10^{4} N C^{- 1} . (a)$ , we see that the absolute values of the charges on beads $1$ and $2$ must be in the ratio of $5$ to $4$ . This checks with the $180 °$ value from the $E_{x}$ graph, which further confirms our belief that bead $1$ is positively charged. In fact, the $180 °$ value from the $E_{x}$ graph allows us to solve for its charge.

$q_{1} = 4 {πε}_{0} r^{2} E = 4 π (8.854 \times 10^{- 12} \frac{C^{2}}{N m^{2}}) {(0.40 m)}^{2} (5.0 \times 10^{4} \frac{N}{C}) = + 0.88 \times 10^{- 6} C$

$(c)$ Similarly, the $0 °$ value from the $E_{y}$ graph allows us to solve for the charge of bead $2$ :

$E_{y s} = - 9.0 \times 10^{4} N C^{- 1} . (a)$

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