m=6.8•10⁻³ kg, q=0.161•10⁻⁶C, A=0.0155 m², α=30°, Q=?
Let’s find the projections of the forces acting on the plastic ball:
x-projections: Tsinα =F ……. (1)
y-projections: Tcosα = mg ……(2)
Divide (1) by (2)
Tsinα /Tcosα=F/mg,
tanα =F/mg…………(3)
F=qE=qσ/ε₀=q•Q/A ε₀ …(4)
Substitute (4) in (3)
tanα = qQ/A• ε₀•m•g,
Q= A•ε₀•mg• tanα/q=
=0.0155•8.85•10⁻¹²•6.8•10⁻³•9.8•0.577/0.161•10⁻⁶=
= 3.28•10⁻⁸ C
A small plastic ball with a mass of 6.80 10-3 kg and with a charge of +0.161 µC is suspended from an insulating thread and hangs between the plates of a capacitor (see the drawing). The ball is in equilibrium, with the thread making an angle of 30.0° with respect to the vertical. The area of each plate is 0.0155 m2. What is the magnitude of the charge on each plate?
1 answer