Uniform Spherical Charge Distribution Problem - Please Help?

[匿名]

2017-02-13 6:27 am

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.

Part A
Find the total charge contained in the charge distribution.
Express your answer in terms of some or all of the variables r, R, Q, and appropriate constants.

Part B
Obtain an expression for the electric field in the region r≥R.
Express your answer in terms of some or all of the variables r, R, Q, and appropriate constants.

Part C
Obtain an expression for the electric field in the region r≤R.
Express your answer in terms of some or all of the variables r, R, Q, and appropriate constants.

Part D
Find the value of r at which the electric field is maximum.
Express your answer in terms of some or all of the variables r, R, Q, and appropriate constants.

Part E
Find the value of that maximum field.
Express your answer in terms of some or all of the variables r, R, Q, and appropriate constants.

回答 (1)

NCS

2017-02-13 8:47 am

✔ 最佳答案

A) I think we can consider an infinitely thin shell having thickness dr and surface area 4πr². The charge on that shell is
dQ = ρ(r)*A*dr = ρ0(1 - r/R)(4πr²)dr = 4π*ρ0(r² - r³/R)dr
which when integrated from 0 to r is
total charge = 4π*ρ0(r³/3 + r^4/(4R))
and when r = R our total charge is
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3
and after substituting ρ0 = 3Q / πR³ we have
total charge = Q ◄

B) E = kQ/d²
since the distribution is symmetric spherically

C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 - r/R)dr
so
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is
and after substituting for ρ0 is
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4))
which could be expressed other ways.

D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that
r = R for a min/max (and we know it's a max since r = 0 is a min).

E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R²

barring computational error. Let me know if you would please!

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