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Question
Consider a sphere of radius R with charge density distributed as
ρ(r) = kr for r ≤ R
= 0 for r > R
- Find the electric field at all points r.
- Suppose the total charge on the sphere is 2e where e is the electron charge. Where can two protons be embedded such that the force on each of them is zero. Assume that the introduction of the proton does not alter the negative charge distribution.
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Solution
a. The symmetry of the problem suggests that the electric field is radial. For points r < R, consider a spherical Gaussian surface. Then on the surface
`oint E_r * dS = 1/ε_0 int_v ρdv`
`4pir^2 E_r = 1/ε_0 4pik int_0^r r^('3) dr^'`
= `1/ε_0 (4pik)/4 r^4`
∴ `E_r = 1/(4ε_0) kr^2`
`E_r = 1/(4ε_0) kr^2 hatr`
For points r > R, consider a spherical Gaussian surfaces’ of radius r,
`oint E_r * dS = 1/ε_0 int_v ρdv`
`4pir^2 E_r = (4pik)/ε_0 int_0^r r^3 dr^`
= `(4pik)/ε_0 R^4/4`
∴ `E_r = k/(4ε_0) R^4/r^2`
`E_((r)) = (k/(4ε_0)) (R^4/r^2)hatr`
b. The two protons must be on the opposite sides of the centre along a diameter. Suppose the protons are at a distance r from the centre.
Now, `4pi int_0^R kr^('3)dr = 2e`
∴ `(4pik)/4 R^4 = 2e`
∴ `k = (2e)/(piR^4)`
Consider the forces on proton 1. The attractive force due to the charge distribution is
`-eE_r = - e/(4ε_0) kr^2 hatr = - (2e^2)/(4piε_0) r^2/R^4 hatr`
The repulsive force is `e^2/(4piε_0) 1/(2r)^2 hatr`
Net force is `(e^2/(4 piε_0 4r^2) - (2e^2)/(4piε_0) r^2/R^4)hatr`
This is zero such that `e^2/(16 piε_0 4r^2) = (2e^2)/(4piε_0) r^2/R^4`
Or, `r^4 = (4R^4)/32 = R^4/8`
⇒ `r = R/(8)^(1/4)`
Thus, the protons must be at a distance `r = R/root(4)(8)` from the centre.
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