A Faraday ice pail similar to the one you will use is shown to the right. The pr

Question

A Faraday ice pail similar to the one you will use is shown to the right. The principle of its operation relies on the proportionality between the charge on the inner pail and the potential difference between the inner pail and ground. Suppose the relationship between charge q and potential difference V is given by V=2.5 × 1016 q. When a charged wand is inserted into the inner pail and touches it, the electrometer reading is -18.0 volts. (a) How much charge resided on the wand? (b) Before grounding the pail, a second wand is inserted into the ice pail and touches it. The electrometer reading now is +30.0 volts. How much charge resided on the second wand? (c) If the two wands were separated by 4.0 cm before they were discharged onto the ice pail, what magnitude of force did either exert on the other? Was it attractive or repulsive? Assume that no leakage of charge occurs during the experiment and that charge on the wand is same as the charge on the pail. Recall that 1 / (4 pi e0) = 9 × 10^9 Nm^2/C^2

5. The two wands are rubbed together and held 1.00 cm apart. If the charge on the white-faced wand is 1.2 x 10^-10 C, what is the magnitude of the force each wand exerts on the other one? Is the force attractive or repulsive? State your assumptions.

Explanation / Answer

(a)

the formula for potental pint charge is

2.5 x 10^16 q1 = - 18 volts

q1 = - 7.2 x 10^-16 C

(b)

V =  30 volts

2.5 x 10^16 ( q1 + q2 ) = 30 volts

q1 + q2 = 1.2 x 10^-15 C

q2 = 1.2 x 10^-15 - ( - 7.2 x 10^-16 )

     = 1.2 x 10^-15 + 0.72 x 10^-15

     = 1.92 x 10^-15 C

(c)

F = k q1 q2 / r^2

   = 9 x 10^9 * 7.2 x 10^-16 * -1.92 x 10^-15 / ( 4 x 10^-2)^2

   = - 7.77 x 10^-18 N

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5

F = k q1 q2 / r^2

   = 9 x 10^9 * 1.2x 10^-10 *( 1.2 x 10^-10) / ( 1 x 10^-2)^2

   = 1.29 x 10^-6 N

force is repulsive

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