Two 0.22 g pith balls are suspended from the same point by threads 40 cm long. (

Question

Two 0.22 g pith balls are suspended from the same point by threads 40 cm long. (Pith is a light insulating material once used to make helmets worn in tropical climates.) When the balls are given equal charges, they come to rest 18 cm apart, as shown in the figure.
(Figure Image: http://session.masteringphysics.com/problemAsset/1138195/3/Figure_15.28.jpg)

What is the magnitude of the charge on each ball? (Neglect the mass of the thread.)

Explanation / Answer

Given two 0.22 g pith balls are suspended from the same point by threads 40 cm long T = m*g / cos(theta) -->Sum of the forces in the horizontal direction on one of the balls = 0 0 = T*sin(theta) - F_charge -->force by charge = T*sin(theta) = m*g*sin(theta)/cos(theta) = m*g*tan(theta) -->forcebycharge = k*q1*q2/r^2 Since q1 = q1 =q -->F= k*q^2/r^2 = m*g*tan(theta) q = sqrt(m*g*r^2/k*tan(theta)) q = sqrt( .00026 kg * 9.81 m/s^2 * (.18 m)^2 / (9.0*10^9 N/(m^2*C^2)) * tan (asin(9cm/30cm)) -->q = 5.37*10^-8 C

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