Question

A spherically symmetric charge distribution has a charge density given by ρ = a/r , where a is constant. Find the electric field within the charge distribution as a function of r. Note: The volume element dV for a spherical shell of radius r and thickness dr is equal to 4πr2dr. (Use the following as necessary: a, r, and ε0. Consider that a is positive.)

Answer 1

Answer:

E = a/2ε۪

Explanation:

A spherically symmetric charge distribution has a charge density given by ρ = a/r , where a is constant. Find the electric field within the charge distribution as a function of r. Note: The volume element dV for a spherical shell of radius r and thickness dr is equal to 4πr2dr. (Use the following as necessary: a, r, and ε0. Consider that a is positive.)

the charge on a spherical ball is the following

dQ = (charge density) × (surface area) × dr

dQ = ρ(r)4πr²dr

integrating the both sides, we have

∫ dQ = ∫ (a/r)4πr²dr

∫ dQ = 4πa ∫ rdr

taking between limit r=r and r=0

Q(r) = 2πar² - 2πa0²

Q = 2πar² (= total charge bound by a spherical surface of radius r)

(Flux out of surface) = (charge bound by surface)/ε۪  

Gauss law

(Surface area of sphere) × E = Q/ε۪  

4πr²E = 2πar²/ε۪  

E = a/2ε۪

Answer 2

Charge dQ on a shell thickness dr is given by 

dQ = (charge density) × (surface area) × dr 

dQ = ρ(r)4πr²dr 

∫ dQ = ∫ (a/r)4πr²dr 

∫ dQ = 4πa ∫ rdr 

Q(r) = 2πar² - 2πa0² 

Q = 2πar² (= total charge bound by a spherical surface of radius r) 

Gauss's Law states: 

(Flux out of surface) = (charge bound by surface)/ε۪ 

(Surface area of sphere) × E = Q/ε۪ 

4πr²E = 2πar²/ε۪ 

E = a/2ε۪


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