Answer:
a) Q = π b a⁴ , b) E = k π b a⁴ / r² , E = kπ b a² , E = k π b r² ,
d) V = - k π b a⁴ / r , V = - k π b r³ / 3
Explanation:
a) the charge density is defined by
ρ = dQ / dV
dQ = ρ dV
the volume differential is the area of the sphere by the change in radius
dV = (4π r²) dr
we integrate to enter the charge
Q = ∫ ρ (4π r²) dr
we replace and integrate
Q = 4π b ∫ r r² dr
Q = 4π b r⁴ / 4
we evaluate between the integration limits, lower r = 0 and higher r = a
Q = π b a⁴
We define a Gaussian surface in spherical shape, in this case the field line is parallel to the radii of the sphere, so the scalar product is reduced to the ordinary product
The area of a sphere is
A = 4π r²
E 4π r² = qint /ε₀
E = qint / (4πε₀) 1 / r²
k = 1 / 4π ε₀
we must find the charge inside the gaussian surface
b) Calculate the electric field for different points, for this we use Gauss's law
i) r> a
Ф = E . dA = qint /ε₀
qint = π b a4
whereby the electric field is
E = k π b a⁴ / r²
ii) r = a
E = kπ b a²
iii) 0 <r <a
in this case the charge inside the gaussian surface is
qint = π b r⁴
the electric field is
E = k π b r⁴ / r²
E = k π b r²
c) see attached
d) The electric potential is
ΔV = - ∫ E .dr
we have two regions
i) r> a
ΔV = - ∫ (k pi b a⁴ / r²) dr
ΔV = - k π b a⁴ ∫ dr / r²
ΔV = - k π b a⁴ (1 / r)
we evaluate for the lower limit r = r and the upper limit r = ∞, we make the potential zero for infinite distance
V = - k π b a⁴ / r
ii) in the region
0 <r <a
ΔV = - ∫ (k π b r²) dr
ΔV = -k π b r³ / 3
in this case we make that for r = 0 the potential is zero
V = - k π b r³ / 3
