Question

Imagine to neutralize instantaneously all the electrons of a solid sphere of 1cm in diameter.
The remaining positive charge q produces an explosion of the matter due to electrostatic repulsion.
i) Calculate an expression for the energy release, in terms of this remaining charge q.
ii) Estimate a numerical value for this energy, expressed in units of Mtonnes of TNT if the chosen material is Carbon.
Hint. Calculate the work done to assemble a uniformly charged sphere from charges which are originally at infinity...

Answer 1

Expression for the electrostatic energy in terms of remaining charge q is given by,

E = 1/(4 pi epsilon not) {3q^2/5R} ...(A)

Here R is the radius of sphere.

(b) The remaining charge q is calculated as = (total protonic charge in carbon)×(total number of carbon atom)×(charge on one proton)

Total number of carbon atom = radius of sphere ÷ atomic radius of carbon

n = (R/a)^3 ...(1)

Here n is the total number of carbon atom and a is the radius of carbon atom or atomic radius of carbon.

Substitute 0.5 Cm for R and 0.7 angstrom for a in equation (1)

n = {0.5×10^(-2)}÷{0.7×10^(-10)}

= 3.640×10^(23)

Atomic number of carbon = total number of proton

Total protonic charge = total number of proton× charge on one proton

Total protonic charge = {6× 1.6×10^(-19)}

= 9.6×10^(-19)

Now total charge q on the sphere is,

q = n×9.6×10^(-19)

= {3.640×10^(23)}{9.6×10^(-19)}

= 3.50×10^(5) Coulomb

Substitute 9×10^(9) for 1/4 pi epsilon not, 3.50×10^(5) for q and 0.5 ×10^(-2) for R in equation (A).

E = {9×10^(9)} (3){3.50×10^(5)}^2÷(5){0.5×10^(-2)}

= 13.2×10^(22) Joules

Conversion from Joules to Mtonnes.

1 Mtonnes = 4.184×10^(15) Joules

E = 13.2×10^(22)÷4.184×10^(15)

= 31.6×10^(6)

= 3.16×10^(7) Mtonnes