Question

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?

Answer 1

Upon contact and separation, each of the four identical metallic spheres will have a charge of -1.95 µC, corresponding to approximately 12,180 excess electrons.

Step-by-step explanation:

When four identical metallic spheres with charges of 2.2 µC (microcoulombs), 4.2 µC, −7.4 µC, and −6.8 µC are brought into contact and then separated, the resulting charge on each sphere is found by calculating the average of the total charge. This is a demonstration of the conservation of charge, which states that the total charge in an isolated system remains constant. To find the average, we add up all the charges and divide by the number of spheres:

Total charge = 2.2 µC + 4.2 µC − 7.4 µC − 6.8 µC = −7.8 µC

To find the average charge per sphere:

Average charge per sphere = Total charge / Number of spheres = −7.8 µC / 4 = −1.95 µC

Each sphere will have a charge of −1.95 µC after separation. To determine the number of excess or absent electrons, we use the electron's charge, which is approximately 1.602 x 10⁻¹⁹ C (coulombs). The charge on each sphere (in coulombs) divided by the charge of one electron gives:

Number of excess or absent electrons = Resulting charge on each sphere / Charge of one electron = 1.95 x 10⁻¶ C / 1.602 x 10⁻¹⁹ C ≈ 1.218 x 10¹⁴ electrons

This corresponds to approximately 12,180 excess electrons for each sphere, implying that each sphere is negatively charged.

Answer 2

Part a)

charge on each sphere is -1.95 micro coulomb

Part b)

For first sphere

`N = 2.6 * 10^(13)` excess charge

For second sphere

`N = 3.8 * 10^(13)` excess charge

For third sphere

`N = 3.4 * 10^(13)` absent charge

For third sphere

`N = 3.03 * 10^(13)` absent charge

Step-by-step 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

`q = (Q_1 + Q_2 + Q_3 + Q_4)/(4)`

`q = (2.2 \mu C + 4.2 \mu C - 7.4 \mu C - 6.8 \mu C)/(4)`

`q = -1.95 \mu C`

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

`N = (4.15* 10^(-19))/(1.6 * 10^(-19))`

`N = 2.6 * 10^(13)`

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

`N = (6.15* 10^(-19))/(1.6 * 10^(-19))`

`N = 3.8 * 10^(13)`

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

`N = (5.45* 10^(-19))/(1.6 * 10^(-19))`

`N = 3.4 * 10^(13)`

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

`N = (4.85* 10^(-19))/(1.6 * 10^(-19))`

`N = 3.03 * 10^(13)`