Answer:
electric field is k [tex]r^{2}[/tex] / ε × 4
energy is πk² [tex]R^{6}[/tex] /48ε
Explanation:
given data
radius = R
charge density P(r) = kr
r < R
to find out
field of the sphere and energy
solution
we know that P(r) = kr
here k is constant
we divide the sphere in many number of parts and we consider element thickness is dr so voulme will be 4πr²dr
so charge will be
dq = P ( 4πr²)dr
dq = kr ( 4πr²)dr
take integrate both side
∫dq = 4πk ∫r³ dr
q = 4πk [tex]r^{4}[/tex] / 4
q = πk [tex]r^{4}[/tex]
so we apply here gauss law
E×A = q / ε
electric field = πk [tex]r^{4}[/tex] / ε × 4πr²
electric field = πk [tex]r^{4}[/tex] / ε × 4πr²
so electric field = k [tex]r^{2}[/tex] / ε × 4
and
in 2nd part we know
U, energy = energy density × volume
U, energy = (1/2 × ε × E²) × (4πr² )
U energy = (1/2 × ε × (kr²/4ε)²) × (4πr²)
U, energy = πk²[tex]r^{6}[/tex] / 8ε
take integrate both side with 0 to U and 0 to R
[tex]\int_{0}^{U}[/tex]dU = πk²/8ε [tex]\int_{0}^{R}[/tex] [tex]r^{6}[/tex] dr
U, energy = πk²/8ε × [tex]R^{6}[/tex] /6
so energy = πk² [tex]R^{6}[/tex] /48ε
