Question
Mon April 19, 2010 By: Priyanka R

Please solve this sum

Expert Reply
Tue April 20, 2010

Each pith ball will be 0.2/2 = 0.1 m from the vertical centre line.

The vertical component of tension in the string is equal to the weight of a pith ball.

Ty = Tsinθ, where θ is the angle made by string with horizontal.

Hypotenuse = Length of the string = 0.5 m

Adjacent side = 0.1 m

Opposite side = (0.52-0.12)1/2 = 0.49

Ty = Tsinθ = W = 10-3x10 N

T = (10-3x10)/(0.49/0.5) = 10.20x10-3

The horizontal component of tension is the electrostatic force of repulsion,

kq2/r2 = Tx = Tcosθ = 10.20x10-3 (0.1/0.5)

q2 = (10.20x10-3 )(0.1/0.5)(0.22)/(9x109) = 0.009x10-12

q = 0.09 x 10-6 C

q = 90 nC.

Regards,

Team,

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