Two small pith balls, each of mass m = 14.2 g, are suspended from the ceiling of

Question

Two small pith balls, each of mass m = 14.2 g, are suspended from the ceiling of the physics lab by 0.9 m long fine strings and are not moving. If the angle which each string makes with the vertical is ? = 15.7°, and the charges on the two balls are equal, what is the magnitude of that charge (in µC)? (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.)

Two small pith balls, each of mass m = 14.2 g, are suspended from the ceiling of the physics lab by 0.9 m long fine strings and are not moving. If the angle which each string makes with the vertical is ? = 15.7°, and the charges on the two balls are equal, what is the magnitude of that charge (in µC)? (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.)

Explanation / Answer

here

Fy = m *g
Fx = kq^2 / (2*x)^2
x = L sin (theta)............................(1)

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 =4 * x^2 *m *g* tan / k

q= 2 *x * sqrt (m g tan /k)

then from 1 equation

q = 2 * L*sin(theta) * sqrt( m* g * tan(theta) / k)

q = 2 * 0.9 * sin(15.7) * sqrt( 0.0142 * 9.8 * tan(15.7) / 9 * 10^9)

q = 1.015 * 10^-6 C

so the charge on the ball is 1.015 uC

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