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Question

3. [P15-7] Oppositely charged parallel plates are separated by 5.33 mm. A potential difference of 600 V exists between the plates. (a) What is the magnitude of the electric field between the plates? (b) What is the magnitude of the force on an electron between the plates? (c) How much work must be done on the electron to move it to the negative plate if it is initially positioned 2.90 mm from the positive plate?



4. [A15-1b] Rub an inflated balloon with a piece of wool cloth and place the balloon near a fine stream of water falling from a faucet. The stream of water will deflect toward the balloon. Why? Vary the distance between the balloon and the stream, and observe the displacement of the water stream for different distances. What is the relationship between the displacement of the stream and the distance of separation?

Answer 1

3a) for a constant field (as you have with "infinite" parallel plates)
E-field = potential difference / distance

b) Force on the electron = charge * E-field
= fundamental charge * potential difference / distance

c) Work = force * distance
= fundamental charge * potential distance * distance moved / total distance

They give you the total potential difference, total distance, distance moved. Look up the fundamental charge on the electron. Plugnchug

4) Water doesn't conduct electricity, but the mollecules are polar, so you can induce a slight electrostatic dipole moment in the water by applying an E-field. If the E-field varies (as it would for a balloon), this can result in a net force on the water drops.

To put it in precise vector calculus terms, the force will be proportional to the dot product of the induced dipole moment with the gradient of the E-field.

Answer 2

3. a) E=V/d
E=600/(5.33 E-3)=1.13E+5 N/C

b) F=Eq
F=1.13E+5 x 1.602 E-19=1.81 E-14N

c)W=Fd'
W=F (d-a)
W=1.81 E-14N ( 5.33E-3 - 2.90E-3) =4.40 E-17 Joules

4. Hint: Water molecule is polarized.

LOL! Bekki was here 6 hours earlier.
Thumb up to her!