SOLUTION OF POISSON'S EQUATION IN INSULATED SPHERE

TO FIND THE ELECTRIC POTENTIAL INSIDE THE CHARGED INSULATING SPHERE

For this purpose most suitable method is solving Poisson’s equations, because inside the insulating charged sphere (ICS) there is a constant charge density and it is also an easy method.

The Poisson’s equation, ∆ ²V= -  Ω/€₀  Where, Ω is the volume charge density. In ICS potential is the function of polar distance(r). So we can change Laplacian of V, using vector differential rule.

∆²V= d²V/dr² + (2/r)dV/dr = -Ω/€₀      or

(r) d²V/dr² + 2dV/dr = -rΩ/€₀

To solve this differential equation by Integral series.

Let, V = c₀ + c₁r + c₂r² + c₃r³ + c₄r⁴+ c₅r⁵ + …………………………..

Then,

dV/dr = c₁ + 2c₂r + 3c₃r² + 4c₄r³ + 5c₅r⁴ + ………………………………

d²V/dr² = 2c₂ + 6c₃r + 12c₄r³ + 20c₅r⁴ + …………………………………….

Therefore the equation,

(r) d²V/dr² + 2dV/dr = -rΩ/€₀

Will be equals to

[2c₂r + 6c₃r² + 12c₄r³ + 20c₅r⁴ + ………..] + [2c₁ + 4c₂r + 6c₃r² + 8c₄r³ + 10c₅r⁴ + …………….]

To equating the coefficient of r, we get c₂ = -Ω/6€₀

And also equating the coefficients of     constant, r, r²,r³,r⁴ to zero

We get,

      c₁ = c₃ = c₄ = 0    and so on.

Therefore the required solution

V = c0 - r 2Ω/6€0

In ICS,

  Ω =  3Q/(4πr³)

At the surface of the sphere, V is V₀ = (Q /[4πR€₀]) and r = R;

V₀ = c₀ – R²Ω/6€₀ = c₀ – QR² /[8πR³€₀]

ie, (Q /[4πR€₀]) = c₀ – QR² /[8πR³€₀]    and c₀ = 3Q /[8πR€₀]

Therefore the potential inside the insulated sphere is equals to,

V = Q /[8πR€0] {3 – (r2/R2)}

It derived by myself therefore i don't sure that this is the solution, so kindly forgive to me.