f. Drive the Poisson's equation and give the conditions under which it reduces to the Laplace's equation. [4 Marks] g. Show that the potential is constant inside an enclosure completely surrounded by conducting material , provided there is no charge within the enclosure. [4 Marks] h. Find the energy of uniformly charged spherical shell of total charge q and radius R. 3 Marks]

f. To derive the Poisson's equation, we start with Gauss's law, which states that the total electric flux through a closed surface is equal to the total charge enclosed divided by the permittivity of the medium. Mathematically, this can be expressed as:<br /><br />∮E⋅dA=Qenclosedε0<br /><br />where E is the electric field, dA is the differential area, and Qenclosed is the total charge enclosed by the closed surface. Now, let's consider a volume V enclosed by the closed surface. The total charge enclosed Qenclosed can be expressed as the charge density ρ times the volume V:<br /><br />Qenclosed=∫∫∫ρdV=∫∫∫∇⋅EdV<br /><br />Using Gauss's law, we can rewrite the above equation as:<br /><br />∫∫∫∇⋅EdV=∮E⋅dAε0<br /><br />Now, let's consider the Laplace's equation, which states that the sum of the second partial derivatives of a scalar function φ (such as electric potential) with respect to the coordinates is zero:<br /><br />∇2φ=0<br /><br />This equation describes the distribution of electric potential in a region where there are no charges and no sources of the electric field. To derive the Poisson's equation from the Laplace's equation, we introduce a charge density ρ and a charge-free dielectric medium characterized by a permittivity ε. The electric field E and the electric potential φ are related by the following relationships:<br /><br />∇×E=0∇⋅E=ρε<br /><br />∇φ=-E<br /><br />Taking the divergence of the first equation and using the second equation, we get:<br /><br />∇⋅(∇×E)=∇⋅(−Eε)=0<br /><br />which simplifies to:<br /><br />∇2φ=−1ε∇⋅ρ<br /><br />This is the Poisson's equation, which relates the electric potential φ to the charge density ρ and the permittivity ε of the medium. The conditions under which the Poisson's equation reduces to the Laplace's equation are when there is no charge density (ρ=0) in the region, and the medium is homogeneous and isotropic (ε is constant).<br /><br />g. To show that the potential is constant inside an enclosure completely surrounded by conducting material, we need to consider the behavior of the electric field near the surface of a conductor. According to electrostatics, the electric field inside a conductor is zero. This is because the charges on the surface of the conductor redistribute themselves to cancel any internal electric field. Therefore, the electric field inside the conductor is zero, and the potential is constant throughout the conductor. Since the enclosure is completely surrounded by conducting material, the potential on the surface of the enclosure is the same as the potential inside the enclosure. Therefore, the potential is constant inside the enclosure.<br /><br />h. To find the energy of a uniformly charged spherical shell of total charge q and radius R, we can use the formula for the electric field inside a charged sphere:<br /><br />E=1∣∣∣qR2<br /><br />The energy U stored in the electric field is given by:<br /><br />U=12∫∫∫EdV<br /><br />Since the electric field is spherically symmetric and depends only on the radial distance r from the center of the sphere, we can integrate over the volume of the sphere using spherical coordinates. The volume element in spherical coordinates is dV=r2sinθdrdθdφ. Therefore, the energy U is given by:<br /><br />U=12∫0∞∫0π∫02πE⋅r2sinθdrdθdφ<br /><br />Substituting the expression for E, we get:<br /><br />U=12∫0∞∫0π∫0