(i). A charge of Q coulomb is uniformly distributed over a spherical volume of radius R metre Obtain an expression for the energy of the system.
(ii). What will be the corresponding expression for the energy needed to completely diassemble the planet earth against the gravitational pull amongst its constituent particles? Assume the earth to be a sphere of uniform mass density. calculate the energy, given that the product of the mass and the radius of the earth to be `2.5xx10^(31)kg-m`

Answer

A, B, C

Explanation

Let us consider a shell of the thickness dx at a distance x from the centre of a sphere

The vol. of the shell `=4/3pi[(x+dx)^3-4/3pix^3]`
`=4/3pi[(x+dx)^3-x^3]`
`=4/3pix^3[(1+(dx)/(x))^3-1]`
`=4/3pix^3[1+(3dx)/(x)-1]`
`=4/3pix^3xx(3dx)/(x)=4pix^2dx`
Let `rho` be the charge per unit volume of the sphere
`:.` Charge of the shell `=dq=4pix^2rhodx` ...(i)
Potential at the surface of the sphere of radius x
`=(1)/(4piepsilon_0)xx(rhoxx4/3pix^3)/(x) [:' V=kq/r]`
`:.` Potential at the surface of the sphere of radius x `=(rhox^2)/(3epsilon_0)`
Work done in bringing the charge `dq` on the sphere of radius x
`dW=(rhox^2)/(3epsilon_0)xxdqimpliesdW=(rhox^2)/(3epsilon_0)xx4pix^2rhodx`
Therefore the work done in accumulating the charge Q over a spherical volume of radius R meters
`W=int_0^R(4pirho^2)/(3epsilon_0)x^4dx=(4pirho^2)/(3epsilon_0)[(x^5)/(5)]_0^R=(4pirho^2)/(3epsilon_0)(R^5)/(5)`
`=(4pi)/(3epsilon_0)((Q)/(4//3piR^3))^2(R^5)/(5)=(3Q^2)/(20piepsilon_0R)`
This is also the energy stored in the system.
(b) The above energy calculated is
`E=(3Q^2)/(5xx(4piepsilon_0)R)=(3KQ^2)/(5R)` where `K=(1)/(4piepsilon_0)`
`:.` `E=(3GM^2)/(5R)` [ `:'` Q is replaced by M]
But `g=(GM)/(R^2)impliesgMR=(GM^2)/(R)`
`F=(Kq_1q_2)/(r^2)`, `F=(Gm_1m_2)/(r^2)`
`:.` `E=3/5gMR=3/5xx10xx2.5xx10^(31)=1.5xx10^(32)J`
(c) During the charging process, let at any instant the spherical conductor has a charge q on its surface.
The potential at the surface `=(1)/(4piepsilon_0)xx(q)/(R)`
Small amount of work done in increasing charge dq more on the surface will be
`dW=(1)/(4piepsilon_0)xxq/Rxxdq`
`:.` Total amount of work done in bringing charge Q on the surface of spherical conductor.
`W=(1)/(4piepsilon_0R)int_0^Qqdq=(1)/(4piepsilon_0R)[(q^2)/(2)]_0^Q=(Q^2)/((8piepsilon_0R))`