Two charged beads are on the plastic ring in part ( a ) of the figure below. Bea

Question

Two charged beads are on the plastic ring in part (a) of the figure below. Bead 2, which is not shown, is fixed in place on the ring, which has radius R = 77.0 cm. Bead 1 is initially at the right side of the ring, at angle = 0°. It is then moved to the left side, at angle = 180°, through the first and second quadrants of the xycoordinate system. Part (b) of the figure gives the x component of the net electric field produced at the origin by the two beads as a function of . Similarly, part (c) of the figure gives the y component.

(a) At what angle is bead 2 located? (Assume that it is located in the lower half of the ring.)
______°

(b) What is the charge of bead 1?

_______C

(c) What is the charge of bead 2?
_________C

Explanation / Answer

(a) When bead 1 is on the +y axis, there is no x component of the net electric field, which implies bead 2 is on the -y axis, so its angle is -90 degree.

(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 Ey at 0 deg and at 90 deg 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 deg value from the Ex graph, which further confirms our belief that bead 1 is positively charged. In fact, the 180 deg value from the Ex graph allows us to solve for its charge

q1 = Ex*r^2/k = 5*10^4*(77*10^-2)^2/(9*10^9) = 3.29*10^-6 C

(c) 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.

Similarly, the 0 deg value from the Ey graph allows us to solve for the charge of bead 2:

q2 = Ey*r^2/k = (-4*10^4)*(77*10^-2)^2/(9*10^9) = -2.63*10^-6 C

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