In the future people will only enjoy one sport: Electrodisc. In this sport you g

Question

In the future people will only enjoy one sport: Electrodisc. In this sport you gain points when you cause metallic discs hovering on a field to exchange charge. You are an Electrodisc player playing the popular four disc variant. The disks have charges of qA = 8.0 µC, qB = 2.0 µC, qC = +5.0 µC, and qD = +12.0 µC.

(a) You bring two disks together and then separate them. You measure the resulting charge of these two disks and find that it is 5.0 µC per disk. Which two disks did you bring together?

A and B

A and C

A and D

B and C

B and D

C and D

(b) You bring three disks together and then separate them. You measure the resulting charge of these three disks and find that it is +5.0 µC per disk. Which three disks did you bring together?

A, B, and C

A, B, and D

A, C, and D

B, C, and D

(c) Given the resulting charge of each disk measured in (b) is +5.0 µC, how many electrons would you need to add to a disk of this charge to electrically neutralize it? 3.12e10 How could you use the charge of the electron to calculate the number of electrons in a certain amount of charge?

Explanation / Answer

(a) First option 'A and B' is the correct answer.

Explnation -

when disc A and disc B are brought togather, we have -

qA + qB = -8 - 2 = -10 uC

so charge on each disc = -10 / 2 = -5 uC

(b) Last option 'B,C and D' is the correct answer.

when these three discs are brough togather.

charge on each disc = (-2 + 5 +12) / 3 = 5 uC

(c) total charge on each disc = 5 uC = 5 x 10^-6 C

charge of an electron = 1.6 x 10^-19 C

SO number of electrons need to be added in each disc = (5 x 10^-6) / (1.6 x 10^-19) = 3.125 x 10^13

= 3.12 x 10^13 C

The charge of an electron = 1.6 x 10^-19 C

When we divide a certain amount of charge by the charge of an electron, we can find out the number of elecrons in that amount of charge.

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