Question

Consider a conducting sphere of radius R carrying a net charge Q.

a). Using Gauss’s law in integral form and the equation |E| = σ/ε₀ for conductors, nd the surface charge density on the sphere. Does your answer match what you expect? b). What is the electrostatic self energy of this sphere?
c). Assuming the sphere has a uniform density ρ, what is the gravitational self energy of the sphere? (That is, what amount of gravitational energy is required/released when the sphere is assembled.) d). Assuming Q = 1 C and the sphere is made of ironρ M = 7870 kg/m3, is there a radius R such that the gravitationalself-energy is just enough to overcome the electrostatic self energy? If so, nd R.

Answer 1

GIven that total charge in the sphere is Q. According to Gauss law if you consider any closed surface the sufrace intergral of electric field over it is given by 1/ε0 times the net charge in the system. It doesnt matter how the charge is distrbuted inside it should be enclosed in a closed surface.

E.ds = (1/ε0) * Q

Consider a spherecal surface of Radius R1. R1>> R. Since the Electric Field from the charge is radiallty outwards the dot product at any infinitisimal area is always E.ds = EdsCos() and here   = 0. For spherical surface E is uniform over the surface it you consider R1>> R So integral is just over the surface and is given

Surface Area of the sphere =4*PI*R1^2

E * 4 *PI*R1^2=Q/ε0

Now Consider R1 = R

E = Q / ( 4*PI*R^2)ε0

Now  Q /( 4 *PI*R^2) is nothing but surface charge density σ because due mutual repulsion and mobility of the charge in the conducting sphere,the charges will distribute at the surface.

So E=σ/ε0 or

σ =E*ε0

Since there is no charge reside inside the conducitng sphere Electric field any point R2<R is zero.

b) Electrostatic self energy is nothng but work needed to be done against the mutual repulsion between the charges to arrange the charges in the given distribution form. And is given by W = V(r)q

q is the test charge or charge you introduce to a given distribution

V(r) is the potential due to the charge at the surface at any distance r.

V(r) is also given by ( Electric field)

So E = Q/(4*PI*R^2*ε0)

V(R)= - Q/(4*PI*R*ε0)

Electrostatic self energy is V(R)*q

C) Garvitational self energy = mc^2 where m is the mass is the sphere

c is the velociy of the light in vaccum

m = density * Volume of the sphere

Gravitational self energy =  ρ*(4/3) *Pi*R^3*C^2

d)Given that Q = 1C and assume the test charge is also 1 C .

we need to find R for which gravitational and electrostatic self energies are equal.

ie,

ρ*(4/3) *Pi*R^3*C^2=Qq/ (4*PI* R*ε0)

ρ*(4/3) *Pi*R^3*C^2=Qq/ (4*PI* R*ε0)

or R^4=Qq/{ (4*PI*ε0)*ρ*(4/3) *Pi*C^2)

Putting values for all you will get

Q = 1C q=1C Pi=3.14 ε0=8.99*10^9 ρ = 7870 , C=3*10^8m/s

R^4= 2.986*10^-33

or R = 7*10^-10m