Four identical metallic spheres with charges of 2.2 µC, 4.2 µC, −7.4 µC, and −6.8 µC are placed on a piece of paper. The paper is lifted on all corners so that the spheres come into contact with each other simultaneously. The paper is then flattened so that the metallic spheres become separated.

(a) What is the resulting charge on each sphere?
(b) How many excess or absent electrons (depending on the sign of your answer to part (a)) correspond to the resulting charge on each sphere?

Question

06-09-2019
Physics

contestada

Four identical metallic spheres with charges of 2.2 µC, 4.2 µC, −7.4 µC, and −6.8 µC are placed on a piece of paper. The paper is lifted on all corners so that the spheres come into contact with each other simultaneously. The paper is then flattened so that the metallic spheres become separated.

(a) What is the resulting charge on each sphere?
(b) How many excess or absent electrons (depending on the sign of your answer to part (a)) correspond to the resulting charge on each sphere?

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Answer:

Part a)

charge on each sphere is -1.95 micro coulomb

Part b)

For first sphere

[tex]N = 2.6 \times 10^{13}[/tex] excess charge

For second sphere

[tex]N = 3.8 \times 10^{13}[/tex] excess charge

For third sphere

[tex]N = 3.4 \times 10^{13}[/tex] absent charge

For third sphere

[tex]N = 3.03 \times 10^{13}[/tex] absent charge

Explanation:

Part a)

Since all the spheres are of identical size so the total charge of the sphere will divide equally on them

So we have

[tex]q = \frac{Q_1 + Q_2 + Q_3 + Q_4}{4}[/tex]

[tex]q = \frac{2.2 \mu C + 4.2 \mu C - 7.4 \mu C - 6.8 \mu C}{4}[/tex]

[tex]q = -1.95 \mu C[/tex]

So charge on each sphere is -1.95 micro coulomb

Part b)

For first sphere

initial charge = 2.2 micro coulomb

final charge = -1.95 micro coulomb

excess charge = -1.95 - 2.2 = -4.15 micro coulomb

Q = Ne

[tex]N = \frac{4.15\times 10^{-19}}{1.6 \times 10^{-19}}[/tex]

[tex]N = 2.6 \times 10^{13}[/tex]

For second sphere

initial charge = 4.2 micro coulomb

final charge = -1.95 micro coulomb

excess charge = -1.95 - 4.2 = -6.15 micro coulomb

Q = Ne

[tex]N = \frac{6.15\times 10^{-19}}{1.6 \times 10^{-19}}[/tex]

[tex]N = 3.8 \times 10^{13}[/tex]

For third sphere

initial charge = -7.4 micro coulomb

final charge = -1.95 micro coulomb

absent charge = -1.95 + 7.4 = 5.45 micro coulomb

Q = Ne

[tex]N = \frac{5.45\times 10^{-19}}{1.6 \times 10^{-19}}[/tex]

[tex]N = 3.4 \times 10^{13}[/tex]

For third sphere

initial charge = -6.8 micro coulomb

final charge = -1.95 micro coulomb

absent charge = -1.95 + 6.8 = 4.85 micro coulomb

Q = Ne

[tex]N = \frac{4.85\times 10^{-19}}{1.6 \times 10^{-19}}[/tex]

[tex]N = 3.03 \times 10^{13}[/tex]