Question

A thin circular sheet of glass with a diameter 5 meters is rubbed with a cloth on one surface and becomes charged uniformly. A chloride ion (a chlorine atom which has gained one extra electron) passes near the glass sheet. When the chloride ion is near the center of the sheet, at a location 0.8 mm from the sheet, it experiences an electric force of 8e-15 N, toward the glass sheet. In this problem, use the value ε₀ = 8.85e-12 C²/(N·m²)

How much charge is on the surface of the glass disk? (give amount, including sign, and correct units)

Answer 1

The diameter of the disk d = 5 m

The radius of the disk is r = 2.5 m

The distance of the ion from the center of the disk is z = 0.8 mm = 0.8*10^-3 m

The force experienced by the ion is F = 8*10^-15 N

The charge on the ion is q = 1.6*10^-19 C

The force experienced by a chargfe of magnitude q in a n electric field of strength E is given by

F = E*q

Where E is the electric field produced by the uniformly charged disk.

The electric field at a distance z from the center of the disk with radius r and uniform charge density is given by the equation

E = k**2*pi [1-z/sqrt( z² + r² )]

Where k = 9*10⁹ Nm²/C² is the Coulomb's constant

is the surface charge density

Therefore

8*10^-15 N = ( 9*10⁹)*(1.6*10^-19 )*()*(2pi)[ 1-(0.8*10^-3 )/sqrt { (0.8*10^-3)²+(2.5)²} ]

= 8.842*10^-7 C/m²

Let Q be the charge enclosed by the disk, then

Q/pi*r² = 8.842*10^-7 C/m²

Q = 8.842*10^-7*3.14*(2.5)²

Q = 17.354*10^-6 C

Q = 17.354 uC

Q = +12.5 μC