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Paraphin [41]
3 years ago
7

A 49.3 g ball of copper has a net charge of 2.0 µc. what fraction of the copper's electrons has been removed? (each copper atom

has 29 protons, and copper has an atomic mass of 63.5.)
Physics
1 answer:
Serggg [28]3 years ago
7 0
First, find how many copper atoms make up the ball: 
 moles of atoms = (49.3 g) / (63.5 g per mol of atoms) = 0.<span>77638</span><span>mol 
</span> # of atoms = (0.77638 mol) (6.02 × 10^23 atoms per mol) = 4.6738*10^23<span> atoms </span>

<span> There is normally one electron for every proton in copper. This means there are normally 29 electrons per atom:
</span> normal # electrons = (4.6738 × 10^23 atoms) (29 electrons per atom) = <span> <span>1.3554</span></span><span>× 10^25 electrons 
</span>
<span> Currently, the charge in the ball is 2.0 µC, which means -2.0 µC worth of electrons have been removed.
</span><span> # removed electrons = (-2.0 µC) / (1.602 × 10^-13 µC per electron) = 1.2484 × 10^13 electrons removed
 
</span><span> # removed electrons / normal # electrons = </span>
<span>(1.2484 × 10^13 electrons removed) / (1.3554 × 10^25 electrons) = 9.21 × 10^-13 </span>

<span> That's 1 / 9.21 × 10^13 </span>
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Gnoma [55]

A) Calculate the resultant electric field strength at the midpoint between the charges.

Qx is the charge at X and Qy is the charge at Y.

E at midpoint = k×Qx/0.25² - k×Qy/0.25²

k = 9×10⁹Nm²C⁻², Qx = 45nC, Qy = 12nC

E = 4752N/C

Well done.

B) Calculate the distance from X at which the electric field strength is zero.

Let D be some point between X and Y for which the net E field is 0.

Let d be the distance from X to D.

Set up the following equation:

E at D = k×Qx/d² - k×Qy/(0.5-d)² = 0

Do some algebra to solve for d:

k×Qx/d² = k×Qy/(0.5-d)²

Qx/d² = Qy/(0.5-d)²

Qx(0.5-d)² = Qyd²

(0.5-d)√Qx = d√Qy

0.5√Qx-d√Qx = d√Qy

d(√Qx+√Qy) = 0.5√Qx

d = (0.5√Qx)/(√Qx+√Qy)

Plug in Qx = 45nC, Qy = 12nC

d ≈ 330mm

C) Calculate the magnitude of the electric field strength at the point P on the diagram below.

First determine the angles of the triangle. The sides of the triangle are 0.3m, 0.4m, and 0.5m, so this is a right triangle where the angle between the 0.3m and 0.4m sides is 90°

∠Y = tan⁻¹(0.4/0.3) = 53.13°

∠X = 90-∠Y = 36.87°

Determine the horizontal component of E at P:

Ex = E from Qx × cos(∠X) - E from Qy × cos(∠Y)

Ex = k×Qx/0.4²×cos(36.87°) - k×Qy/0.3²×cos(53.13°)

Ex = 1305N/C

Determine the vertical component of E at P:

Ey = E from Qx × sin(∠X) - E from Qy × sin(∠Y)

Ey = k×Qx/0.4²×sin(36.87°) - k×Qy/0.3²×sin(53.13°)

Ey = 2479N/C

Use the Pythagorean theorem to determine the magnitude of E at P:

E = √(Ex²+Ey²)

E ≈ 2802N/C

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