Question

A sphere of radius R carries a charge density P(r) = kr, for r < R. (a) What is the field of the sphere, as a function of r? (b) What is the energy of this configuration?

Answer 1

Answer:

electric field is  k $r^{2}$  / ε × 4

energy  is  πk² $R^{6}$   /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 $r^{4}$ / 4

q = πk $r^{4}$

so we apply here gauss law

E×A = q / ε

electric field =  πk $r^{4}$  / ε × 4πr²

electric field =  πk $r^{4}$  / ε × 4πr²

so electric field =  k $r^{2}$  / 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$r^{6}$ / 8ε

take integrate both side with 0 to U and 0 to R

$\int_{0}^{U}$dU = πk²/8ε $\int_{0}^{R}$ $r^{6}$ dr

U, energy =  πk²/8ε × $R^{6}$ /6

so energy =   πk² $R^{6}$   /48ε