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 BA and C    A and DB and CB and DC 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

+3.0 µC

per disk. Which three disks did you bring together?

A, B, and CA, B, and D    A, C, and DB, C, and D

(c) Given the resulting charge of each disk measured in (b) is

+3.0 µC,

how many electrons would you need to add to a disk of this charge to electrically neutralize it?
electrons

Explanation / Answer

Given; qA = -8 C ; qB = -2 C; qC = 5 C and qD = 12 C

(a) As the final charge on each disc becomes 5 C therefore disc B and D must have been brought together so that their average charge become 1/2*(12 - 2) = 5 C

(b) As the charge on all three disc is 3 C hence the combination must have had total charge of 3*3 = 9 C. So disc A, C and D must have been brought together so that the total charge became -8 + 5 + 12 = 9 C

(c) As the charge of one electron is e = -1.6*10-19 C hence the number of electrons (n) needed to make

3 C = 3*10-6 C charged disc neutral would be :

n =   3*10-6/1.6*10-19 = 1.875*1013 electrons

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.