A thin circular sheet of glass of diameter 5 meters is rubbed with a cloth on on

Question

A thin circular sheet of glass of 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 th sheet, at a location 0.9 mm from the sheet, it experiences an electric force of 8e-15 N, toward the glass sheet. In this problem, use the value g = 8.85e-12 C^N^m^) It will be useful to you to draw a diagram on paper, showing field vectors, force vectors, and charges, before answering the following questions about this situation. (a) Which of the following statements about this situation are correct? Select all that apply, The electric field of the glass sheet is equal to the electric field of the chloride ion. The force on the chloride ion is equal to the electric field of the glass sheet. The net electric field at the location of the chloride ion is zero. The charged disk is the source of the electric field that causes the force on the chloride ion. The electric field that acts on the chloride ion is due to the charge on the glass sheet and to the charge on the chloride ion. (b) In addition to an exact formula for the electric field of a disk, the textbook derives two approximate formulas. In the current situation we want an answer that is correct to 3 significant figures. Which of the following is correct? 0 R >> z, so it is adequate to use the most approximate formula here. z is nearly equal to R, so we have to use the exact formula, z

Explanation / Answer

a) The only statement correct is this,

The charged disk is the source of the electric field that causes the force on the chloride ion.

b) Here you only need to look at your R and z values. R = 2.5m and z = 9e-3 m Therefore, your z <<R. You can use an approximation when your z and R values are very different in size.

c) Here you need to use the equation:
F = (-1.6e-19)(Q/(R^2*pi))/(2*Eo)

8e-15 =(-1.6e-19)(Q/(2.5^2*3.14))/(2*8.85e-12)

Q=-1.08*10^17 C

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