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.



