Question

A positive charge distribution exists within a nonconducting spherical region of radius a. The volume charge density rho is ​not ​uniform but varies with the distance r from the center of the spherical charge distribution, according to the relationship rho = br for 0 ​<​ r ​<​ a, where b is a positive constant, and rho = 0, for r > a.a. In terms of B, a, and fundamental constants, determine the total charge Q in the spherical region.
b. In terms of B, r, a, and fundamental constants, determine the magnitude of the electric field at a point a distance r from the center of the spherical charge distribution for each of the following cases.
i. r > a
ii. r = a
iii. 0 < r < a

c. Quantitatively graph the electric field vs. r for 0 < r , 3a. Include labels and values on the axes.

d. In terms of B, a, and fundamental constants, determine the electric potential at a point a distance r from the center of the spherical charge distribution for each of the following cases:
i. r > a
ii. 0 < r < a

Answer 1

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