Chapter 18, Problem 04 GO Four identical metal spheres have charges of q~-8.0 uc

Question

Chapter 18, Problem 04 GO Four identical metal spheres have charges of q~-8.0 uc, q 2.0 uc, ac= +5.0 pC, and qD=+ 12.0 nc. (a) Two of the spheres are brought together so they touch, and then they are separated. Which spheres are they, if the final charge on each one is +5.0 uC? (b) In a similar manner, which three spheres are brought together and then separated, if the final charge on each of the three is +3.0 ? (c) The final charge on each of the three separated spheres in part (b) is +3.0 uC. How many electrons would have to be added to one of these spheres to make it electrically neutral? is +3.0 yC. How many e N Number Units the tolerance is +/-396 Click if you would like to Show Work for this question: Open Show Work

Explanation / Answer

A ) Let X and y be the intial charges on the spheres.

x + y = 2 * final charge = 2 x 5µC

x+y = +10µC

so x and y are qB and qD

B) x+y+z=3*(+3µC) = +9µC

-8+5+12 = +9

So the answer is qA , qC and qD

C) a charge of -3µC is needed to neutralize one sphere

Number of electrons n = -3 x10-6 / charge of electron e

N = -3x10-6/-1.6x10-19

= 1.875 x1013 electrons are needed

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F = 8.2 N

Let q and q’ be two charges at a distance r the electrostatic force F = kqq’/r2

If the distance is reduced from r to r/9 the the force F’ = kqq’/(r/9)2

F’ = 81 x kqq’/r2 = 81 x F= 664.2 N

Number of electrons in one molecule = 10

Number of electrons in one mole = Avogadro number x 10

=6.023x1023x10

= 6.023x1024

Charge of one mole of electron = 6.023x1024 x (-1.6x10-19)

= -9.6368x105 C

Since density of water is 1 kg/liter

Mass of one liter water is 1 kg = 1000 g

Number of moles in one liter water = 1000/18 =

So charge of electrons in one liter water is = -9.6368x105 x 1000/18 = - 535.4x105 C = -5.354x107 C

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