(a) Calculate the charge on the bead. Q b = (b) If the mass of the bead were tri

Question

(a) Calculate the charge on the bead.

Qb =

(b) If the mass of the bead were tripled (m' =3m), how far above the origin wouldit now float atequilibrium?

yb' =

(c) What is this new charge?

Qc =

QuestionDetails: A rigid, insulating fiber runs along a portion of they-axis; the fiber isnot free to move. Gravity actsdownward (g = 9.81 m/s2).A chargeQa = -5 µC is fixed to the fiber at theorigin. A bead with a hole drilled through its center is slippedover the fiber andis free to move along the fiber without friction.The mass of the bead is m = 150 gand its charge isQb. At equilibrium, the bead floats adistanceyb = 12 cm above the origin.

(a) Calculate the charge on the bead.

Qb =

(b) If the mass of the bead were tripled (m' =3m), how far above the origin wouldit now float atequilibrium?

yb' =

The bead (with original mass m) is removed and itscharge altered to Qc. Whenplaced on the fiberfrom below, it floats at equilibrium point yc =-11 cmbelow the origin.

(c) What is this new charge?

Qc =

Explanation / Answer

g = 9.81 m/s2, Qa = -5µC, m = 150 g, yb = 12 cm a) find Qb. At equilibrium, mg =kQaQb/yb2 Qb = mgyb2/(kQa)= -4.71*10-7 C b) m' = 3m m'g =kQaQb/yb'2m/m' =(yb'/yb)2 yb' = yb*(m/m') = 6.93 cm c) yc = -11 cm At equilibrium, mg +kQaQc/yc2 = 0 Qc =-mgyc2/(kQa) =+3.95*10-7 C

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