A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows:
ρ(r)=ρ0(1−r/R) for r≤R
ρ(r)=0 for r≥R
where ρ0=3Q/πR3 is a positive constant.

Express your answer in terms of some or all of the variables r, R, Q, and appropriate constants.
Obtain an expression for the electric field in the region r≥R.
Obtain an expression for the electric field in the region r≤R.
Find the value of r at which the electric field is maximum.
Find the value of that maximum field.

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)

codiepienagoya codiepienagoya · Answer 2 · 2021-10-06T18:53:40+02:00

For question 1:

For the region [tex]r \geq R[/tex], an expression for the electric field can be calculated by using the Gauss law:

Use gauss law of electrostatics, the charge electric flux through the Gaussian surface is,

[tex]\int \vec{E} \cdot d\vec{A}=\frac{q_{enc}}{\varepsilon_{0} }\\\\E(4\pi r^2)=\frac{Q}{\varepsilon_0}\ \ \ \ \ \ \ \ \ \ \ \ \ (Since\ \ q_{enc}=Q )[/tex]

Here, Q is the enclosed charge and [tex]\varepsilon_0[/tex] , is the permeability.

[tex]E=\frac{1}{4\pi \varepsilon_0 } \frac{Q}{r^2}[/tex]

Therefore, the required electric field is [tex]\frac{Q}{4\pi\varepsilon_0 r^2 }[/tex]

For question 2:

For Gaussian sphere of radius [tex]r \leq R[/tex], the charge enclosed within a spherical shell is,

[tex]dq =\rho_0(1-\frac{r}{R}) 4\pi r^2 dr[/tex]

Hence, the total charge enclosed with in the radius r is as follows,

[tex]q_{enc}=\int^{r}_{0} \rho_{0} (1-\frac{r}{R}) 4\pi r^2 \ dr\\\\[/tex]

[tex]=4\pi \rho_0 |\frac{r^3}{3}-\frac{r^4}{4R}|^4_{0}\\\\=\frac{12 Q}{R^3}(\frac{r^3}{3}-\frac{r^3}{4R})\\\\=\frac{Qr^3}{R^3}(4-\frac{3r}{R})\\\\[/tex]

Use Gauss's law for the Gaussian sphere of radius r,

[tex]\int \vec{E}\cdot dA=\frac{q_{enc}}{\varepsilon_0}\\\\E(4\pi r^2)=\frac{Qr^3}{\varepsilon_0 R^3}(4-\frac{3r}{R})\\\\E=\frac{Qr}{4\pi \varepsilon_0 R^3}(4-\frac{3r}{R})\\\\[/tex]

Therefore, the required electric field at [tex]r\leq R[/tex] is [tex]\frac{Qr}{4\pi \varepsilon_0 R^3}(4-\frac{3r}{R})\\\\[/tex]

For question 3:

The change in electric filed with respect to distance is zero when the electric field has a maximum value.

[tex]\frac{dE}{dr}= 0\\\\\frac{d(\frac{Qr}{4\pi \varepsilon_0 R^3}(4-\frac{3r}{R}))}{dr}=0\\\\(\frac{Q}{4\pi \varepsilon_0 R^3})\frac{d(4r-\frac{3r^2}{R})}{dr}=0\\\\[/tex]

Here, [tex](\frac{Q}{4\pi \varepsilon_0 R^3})[/tex] is not equal to zero. Hence,

[tex]\frac{d(4r-\frac{3r^2}{R})}{dr}=0\\\\4-\frac{6r}{R}=0\\\\r=\frac{4R}{6}=\frac{2R}{3}[/tex]

Therefore, the required value of [tex]r \ \ is\ \ \frac{2R}{3}\\\\[/tex]

For question 4:

Maximum value of the electric filed is as follows

[tex]E= \frac{Q}{4\pi \varepsilon_0 R^3} (4-\frac{3r}{R})\\\\[/tex]

[tex]= \frac{Q(\frac{2R}{3})}{4\pi \varepsilon_0 R^3} (4- \frac{3(\frac{2R}{3})}{R})\\\\=\frac{Q}{3 \pi \varepsilon_0 R^2}[/tex]

Therefore, the required maximum value of the electric field is [tex]\frac{Q}{3 \pi \varepsilon_0 R^2}[/tex]

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