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## Problem: You and your friend are discussing last week’s recitation problem, in which an electron falling down a tunnel through a uniformly c

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Problem: You and your friend are discussing last week’s recitation problem, in which an electron falling down a tunnel through a uniformly charged earth exhibits simple harmonic motion. Instead of being uniformly charged, imagine that the Earth is given a spherically symmetric charge density ⍴(r)=b/r, where b is a constant. If an electron is released from rest inside this tunnel at some initial distance from the center, find an expression for the acceleration of the electron inside the tunnel (ignore gravity). If the electron is released from rest at the surface, what value of b will ensure that the electron reaches the center of the Earth in 12 seconds? ………………………………………………………………………………………

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2 months
2021-07-15T17:57:52+00:00
2021-07-15T17:57:52+00:00 1 Answers
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## Answers ( )

Answer:a. a = eb/2ε₀m b. b = 8.91 × 10⁻¹⁸ kg/V

Explanation:a.Using Gauss’ lawε₀∫E.dA = q where E = electric field and q = charge enclosed.

Now , the charge has a spherical symmetric charge density ⍴(r) = b/r, the total charge enclosed at distance r is given by

q = ∫∫∫ρ(r)r²sinθdrdθdΦ

q = ∫∫∫(b/r)r²sinθdrdθdΦ

q = ∫∫∫brsinθdrdθdΦ

we integrate r from 0 to r, θ from 0 to π and Φ from 0 to 2π

q = ∫∫br[∫sinθdθ]drdθdΦ

q = ∫∫br[-cosθ]drdΦ

q = ∫∫br-[cosπ – cos0]drdΦ

q = ∫∫br-[-1 – 1]drdΦ

q = ∫2brdr∫dΦ

q = ∫2brdr[Φ]

q = ∫2brdr[2π – 0]

q = 4πb∫rdr

q = 4πb[r²/2]

q = 4πb[r²/2 – 0]

q = 2πbr²

The let side of the equation is

ε₀∫E.dA = ε₀E∫dA = ε₀E4πr² since there is spherical symmetry

Equating both sides of the equation

ε₀E4πr² = 2πbr²

E = 2πbr²/ε₀4πr²

E = b/2ε₀

Now, the force acting on the electron , F = ma where m = mass of electron and a = acceleration of electron.

F also equal F = eE where e = electron charge and E = electric filed acting on electron

So, eE = ma

a = eE/m

a = eb/2ε₀m

b.I(f the electron is released from rest and is to reach the center of the earth, it covers a distance of the radius of the earth. Using s = ut + 1/2at² where s = radius of earth = R = 6400 km = 6.4 × 10⁶ m , u = initial velocity = 0 m/s. and a = eb/2ε₀m. Substituting these values into s we haveR = 0t + 1/2(eb/2ε₀m)t²

R = ebt²/4ε₀m

making b subject of the formula, we have

b = 4ε₀mR/et²

when t = 12 s and e = 1.609 × 10⁻¹⁹ C, ε₀ = 8.854 × 10⁻¹² F/m and m = 9.109 × 10⁻³¹ kg.

Substituting these values into b we have

b = (4 × 8.854 × 10⁻¹² F/m × 9.109 × 10⁻³¹ kg × 6.4 × 10⁶ m)/(1.609 × 10⁻¹⁹ C × 12²)

b = 2064.67/ 231.696 × 10⁻¹⁸

b = 8.91 × 10⁻¹⁸ Fkg/C

b = 8.91 × 10⁻¹⁸ kg/V