Potential Due to Charged Sphere

Description:

To find Potential due to a non conducting sphere ( where the charge will always be distributed inside the volume of sphere) , take a sphere with total charge inside it be ‘Q’ and the volumetric charge density be ‘ρ’. Let R be the radius of the sphere. We need to calculate potential due to a conducting shell at point ‘P’.

Total Charge = Q

Volumetric Charge density = ρ

We know the relation between volumetric charge density (ρ) and total charge within the volume (Q) :

Q = 4πR³/3 ρ.......(1)

We also know,

dV = -Edr

Case 1

Let the point ‘p’ be outside the charged shell at a distance ‘r’ from its centre and we need to find Potential at that point.

Ɣ > R, V _Outside = 1/4πε₀ Q/r

We know that the Electric field at point outside the non conducting sphere −

E_outside = 1/4πε₀ Q/r²

We also know the relation between electric field (E) and potential (dV)

-dV/dr = 1/4πε₀ Q/r²

dv = -1/4πε₀ Q/r² dr

Integrating both sides of the equation we get,

∫dv = - 1/4πε₀. Q ∫r^-2dr

V = -1/4πε₀. Q r^-1/-1

V = 1/4πε₀. Q r^-1

V_outside = 1/4πε₀. Q r^-1.....(2)

Note −

• When all the charges are in shape of a point charge placed at distance r, the potential is same

  ase that of potential at a point outside the shell.

• The potential goes on increasing As the distance ‘r’ decrease.

Case 2

Let the point ‘p’ be on the surface of the charged shell at a distance ‘r=R’ and we need to find Potential at P.

Ɣon surface = Q/4πε₀r

Putting r = R in equation (1) we get,

V_surface = 1/4πε₀ Q/R

Putting the value of Q from eq (1) −

V_Surface = 4/3πR³ρ/4πε₀R

⇒ V_Surface = R²ρ/3 ε₀.........(3)

Case 3

Let the point ‘p’ be inside the charged shell at a distance ‘r’ from its centre and we need to find Potential at P.

Ɣ < R, V_inside = P(1.5R²-0.5r²)/3ε₀

We know that the Electric field at point inside the non conducting sphere is = R²ρ/3ε₀

.E = -dV/dr

Therefore, dV = - rρ/3 ε₀dr

Integrating the equation within the limit (V_R - V_S)

⇒ V_S∫V_R dV = R∫r - rρ/3ε₀dr

⇒ [V]_Vr^V_S = -ρ/3ε₀[r²/2]_r^R

V_S - V_r = - ρ/3ε₀[R²/2 - r²/2]

From eq (3), V_surface = R²ρ/3ε₀

R²ρ/3ε₀ - V_r = - ρ/3ε₀ R²/2 - r²/2

⇒ V_r = ρ/3ε₀ [R²/2 - r²/2]+ R²ρ/3ε₀

V_r = ρ/3ε₀(1.5 R²-0.5 r²)

Test for V_surface

Put r = R in above equation , we get −

V_r = R²ρ/3ε₀ = 1/4πε₀ = Q/R

Potential at centre

For potential at centre, pur r = 0;

V_center = ρ/3ε₀ * 3R²/2

⇒ V_center = R²ρ/2ε₀

Or,V_center = 3QR²/2*4πε₀R³

V_center = 3Q/2*4πε₀R¹

V_center = (3/2) V_surface