Question 5 Two small pith balls, each of mass m 20 g, are suspended from the cei

Question

Question 5 Two small pith balls, each of mass m 20 g, are suspended from the ceiling of the physics lab by 0.5 m long fine strings and are not moving. If the angle which each string makes with the vertical is = 34.8", and the charges on the two balls are equal, what is the magnitude of that charge (in C)? (Please note that the unit of charge is micro-Coulomb here some browsers might not display the Greek letter mu correctly and show it as an m.) 8.81 10-6 Hint: The tension, the weight, and the Coulomb force must add vectorially to zero. That means that the vertical component of the string tension, T, must equal mg, and the horizontal component must equal the Coulomb force. Check your units! Don't forget to put the answer in C. This is a problem in Statics. And here is an example that reminds you how to deal with these types of problems Submit Answer Incorrect. Tries 3/99 Previous Tries

Explanation / Answer

Fy = mg
Fx = kq^2/(2x)^2
x= L sin
Fx = Ft sin
Fy = Ft cos
Fx/Fy = Ft sin/Ft cos = sin/cos =tan
Fx/Fy = ( kq^2/4x^2 ) / mg =tan
q^2 =4x^2 m g tan / k
q= 2 x sqrt (m g tan /k)

x = 0.5m , m = 0.020kg ,   34.8 , k = 9*10^9N

q= 2 *0.5* sqrt (0.020 *9.81* tan34.8 /9*10^9)

q = 3.892*10^-6C = 3.892µC

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