In the figure two tiny conducting balls of identical mass m and identical charge
ID: 2172146 • Letter: I
Question
In the figure two tiny conducting balls of identical massmand identical chargeqhang from nonconducting threads of lengthL. Assume that ? is so small that tan ? can be replaced by its approximate equal, sin ?. IfL= 130 cm,m= 10 g, andx= 6.4 cm, what is the magnitude ofq, in nanocoulombs?
In the figure two tiny conducting balls of identical massmand identical chargeqhang from nonconducting threads of lengthL. Assume that ? is so small that tan ? can be replaced by its approximate equal, sin ?. IfL= 130 cm,m= 10 g, andx= 6.4 cm, what is the magnitude ofq, in nanocoulombs?Explanation / Answer
Summing forces on the ball in the y directions yields T*cos(?) = mg Summing in the x direction we have FE = k*q^2/x^2 = T*sin(?) Now dividing the 2nd eqn by the first we get sin(?)/cos(?) = k*q^2/(x^2*m*g) or tan(?) ~ sin(?) = k*q^2/(x^2*m*g) but sin(?) = x/2L = Now x/2L = k*q^2/(x^2*m*g) So q = sqrt(x^2*m*g*x/(2L*k)) = sqrt(x^3*m*g/(2L*k)) = sqrt((0.071m)^3*0.013kg*9.8m/s^2/(2*1.50… = 6.35x10^-8C = 63.5nC
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