Answer:
a. a = eb/2ε₀m b. b = 8.91 × 10⁻¹⁸ kg/V
Explanation:
a. Using Gauss' law
ε₀∫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 have
R = 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
