JNTU B.Tech II ELECTROMAGNETIC THEORY AND TRANSMISSION LINES Previous Question Paper (July/August - 2021 - R15)

Gauss's Law states that the total electric flux passing through a closed surface is equal to the charge enclosed divided by the permittivity of free space, expressed as the integral of E dot dS equals Q/ε₀. For an infinite sheet of charge with uniform surface charge density ρs (C/m²), consider a Gaussian pillbox surface extending on both sides of the sheet. Due to symmetry, the electric field E is directed perpendicularly away from the sheet. The flux passes only through the two end caps of the pillbox, each with area A. The total flux is 2EA, and the enclosed charge is ρsA. Equating these, 2EA = ρsA/ε₀, which yields the electric field intensity E = ρs / (2ε₀) in the direction of the unit normal vector.

Given the scalar function V = xy² + x²yz + xyz², the gradient is calculated as the vector sum of partial derivatives with respect to x, y, and z. The partial derivatives are: ∂V/∂x = y² + 2xyz + yz², ∂V/∂y = 2xy + x²z + xz², and ∂V/∂z = x²y + 2xyz. Thus, the gradient is (y² + 2xyz + yz²)a_x + (2xy + x²z + xz²)a_y + (x²y + 2xyz)a_z. The Laplacian is the divergence of the gradient, calculated as the sum of second-order partial derivatives: ∂²V/∂x² + ∂²V/∂y² + ∂²V/∂z². Calculating these: ∂²V/∂x² = 2yz, ∂²V/∂y² = 2x + 2xz, and ∂²V/∂z² = 2xy. Therefore, the Laplacian is 2yz + 2x + 2xz + 2xy.