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2 years ago

A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows:

Physics

1 answer:

Nikitich [7]2 years ago

3 0

Answer:

r ≥ R, E = Q / (4πR²ε₀)

r ≤ R, E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

Maximum at r = ⅔ R

Maximum field of E = Q / (3πε₀R²)

Explanation:

Gauss's law states:

∮E·dA = Q/ε₀

What that means is, if you have electric field vectors E passing through areas dA, the sum of those E vector components perpendicular to the dA areas is equal to the total charge Q divided by the permittivity of space, ε₀.

a) r ≥ R

Here, we're looking at the charge contained by the entire sphere. The surface area of the sphere is 4πR², and the charge it contains is Q. Therefore:

E(4πR²) = Q/ε₀

E = Q / (4πR²ε₀)

b) r ≤ R

This time, we're looking at the charge contained by part of the sphere.

Imagine the sphere is actually an infinite number of shells, like Russian nesting dolls. For any shell of radius r, the charge it contains is:

dq = ρ dV

dq = ρ (4πr²) dr

The total charge contained by the shells from 0 to r is:

q = ∫ dq

q = ∫₀ʳ ρ (4πr²) dr

q = ∫₀ʳ ρ₀ (1 − r/R) (4πr²) dr

q = 4πρ₀ ∫₀ʳ (1 − r/R) (r²) dr

q = 4πρ₀ ∫₀ʳ (r² − r³/R) dr

q = 4πρ₀ (⅓ r³ − ¼ r⁴/R) |₀ʳ

q = 4πρ₀ (⅓ r³ − ¼ r⁴/R)

Since ρ₀ = 3Q/(πR³):

q = 4π (3Q/(πR³)) (⅓ r³ − ¼ r⁴/R)

q = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴)

Therefore:

E(4πr²) = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / ε₀

E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

When E is a maximum, dE/dr is 0.

First, simplify E:

E = 12Q (⅓ (r/R)³ − ¼ (r/R)⁴) / (4πr²ε₀)

E = Q (4 (r³/R³) − 3 (r⁴/R⁴)) / (4πr²ε₀)

E = Q (4 (r/R³) − 3 (r²/R⁴)) / (4πε₀)

Take derivative and set to 0:

dE/dr = Q (4/R³ − 6r/R⁴) / (4πε₀)

0 = Q (4/R³ − 6r/R⁴) / (4πε₀)

0 = 4/R³ − 6r/R⁴

0 = 4R − 6r

r = ⅔R

Evaluating E at r = ⅔R:

E = Q (4 (⅔R / R³) − 3 (⁴/₉R² / R⁴)) / (4πε₀)

E = Q (8 / (3R²) − 4 / (3R²)) / (4πε₀)

E = Q (4 / (3R²)) / (4πε₀)

E = Q (1 / (3R²)) / (πε₀)

E = Q / (3πε₀R²)

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