Fisica III

We make the assumption that bead 2 is in the lower half of the circle, partly because

it would be awkward for bead 1 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 bead 2 is on the –y axis, so its angle is –90°.

(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° and at 90°

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 Ex graph, which further confirms our belief

that bead 1 is positively charged. In fact, the 180° value from the Ex graph allows us to

solve for its charge (using Eq. 22-3):