Aspherically 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.)
Answers
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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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
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