Question

3. A uniformly charged sphere of radius 'a' has a charge density po. Find D everywhere using Gauss's law. Plot the magnitude of D vs R 4. A spherical shell of radius R 1 m has a surface charge density on it of пс and another spherical shell of radius R= 2m has a surface +8 Ps т? пс -6- т2 charge density of ps Find D at R= 3m

Answer 1

D is electric displacement vector inside matter ( dielectrics , not conductors).

Gauss law for insulators is

$\oint D .da = Q_{fenc}$

D = $\epsilon _o$E + P , where P is polarisation the medium.

Q_fenc - is the free charge enclosed inside the Gaussian surface.

charge density = $\rho_o$

Draw a Gaussian sphere of radius R < a , concentric with the center of the sphere.

charge enclosed in side the Gaussian sphere q_en = $\rho_o$ 4/3 $\pi$R³

Like electric field E, D also uniform every where on the surface of the sphere and is normal to the surface.

D* 4$\pi$R² =  $\rho_o$ 4/3 $\pi$R³

D = $\rho_o$R/3 for R<a

for R >= a total charge enclosed inside the Gaussian surface Q_enc = $\rho_o$ 4/3 $\pi$a³

D* 4$\pi$R² =  $\rho_o$ 4/3 $\pi$a³  

D =  $\rho_o$/3a³/R²  

D increases linearly for R<=a to max. $\rho_o$a/3 and then falls with inverse square of R.

- - - - - - + 10 rho*R 3

4. It is not clear how the two shells are placed.