Please solve this sum
Asked by priyankar | 19th Apr, 2010, 09:00: PM
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.
Answered by | 20th Apr, 2010, 09:29: AM
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