A rigid, insulating fiber runs along a portion of the y-axis; the fiber is not f

Question

A rigid, insulating fiber runs along a portion of the y-axis; the fiber is not free to move. Gravity acts downward (g = 9.81 m/s2).A charge Qa = -3 C is fixed to the fiber at the origin. A bead with a hole drilled through its center is slipped over the fiber andis free to move along the fiber without friction. The mass of the bead is m = 210 g and its charge is Qb. At equilibrium, the bead floats a distanceyb = 14 cm above the origin.
Calculate the charge on the bead

I found F=ma to be (0.21)(9.81) then :

F = [k*Qa*Qb]/ (0.14)2

2.0601 = [(8.99x109)*(Qb)*(3x10-6)]/ 0.0196

And found Qb to be 1.497e-6, which is NOT working.

What am I doing wrong??

Explanation / Answer

Sloution: charge fixed on the fiber ,Qa = -3x10-6 C accleration due to gravity , g = 9.8 m/s2 charge on the bead , Qb = ? bead floats a distance ,yb = 14 cm = 0.14 m mass of the bead , m= 210 g = 0.21 kg since , bead with a hole drilled through its center is slipped over the fiber andis free to move along the fiber without friction weight of the bead will equal to electrostatic force acting betweent the charges weight of the bead ,W = mg = (0.21kg)(9.8 m/s2)                                              = 2.058 N The electrostatic force between the charges is ,           F= K Qa Qb /yb2      Here , K is the electrostatic constant = 9x109 Nm2/C2         thus, F = ( -3x10-6 C) (Qb) /(0.14)2   but , W = F    so, 2.058 N = ( 9x109 Nm2/C2   )( -3x10-6 C) (Qb) /(0.14 m)2 thus, Qb = 1.49 x10-6 C    (or ) 1.49 C Don't get confuse this is the correct answer.      you are not doing any wrong                 thus, Qb = 1.49 x10-6 C    (or ) 1.49 C Don't get confuse this is the correct answer.      you are not doing any wrong                

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