JNTU B.Tech III ELECTROMAGNETIC THEORY AND TRANSMISSION LINES Previous Question Paper (September 2021 - R16)

Consider a surface with charge density ρ_s (C/m²) distributed over a surface S. To find the electric field at a point P defined by the position vector r, we divide the surface into infinitesimal elements dS. The charge on each element is dq = ρ_s dS. The electric field dE at point P due to this differential charge is given by Coulomb's Law:

where R is the distance from the source element at r' to the point P, and a_R is the unit vector directed from the source to P. Expressing these in terms of position vectors, R = r - r' and a_R = (r - r') / |r - r'|. Substituting these into the expression:

The total electric field E is the vector sum of these infinitesimal contributions, obtained by integrating over the entire surface S:

Given q₁ = 5 × 10^-9 C at r₁ = (2, 0, 4), q₂ = -2 × 10^-9 C at r₂ = (-3, 0, 5), and test charge Q = 1 × 10^-9 C at r = (1, -3, 7). The vectors from charges to the test point are R₁ = (1-2)a_x + (-3-0)a_y + (7-4)a_z = -a_x - 3a_y + 3a_z (magnitude |R₁| = √19 ≈ 4.36) and R₂ = (1-(-3))a_x + (-3-0)a_y + (7-5)a_z = 4a_x - 3a_y + 2a_z (magnitude |R₂| = √29 ≈ 5.39).

i) Force F = (Q / 4πε₀) * [ (q₁R₁ / |R₁|³) + (q₂R₂ / |R₂|³) ]. Using 1/4πε₀ ≈ 9 × 10⁹, F = 9 * [ (5 * (-a_x - 3a_y + 3a_z) / 19^1.5) + (-2 * (4a_x - 3a_y + 2a_z) / 29^1.5) ] nN. Calculating components: F ≈ (-0.606a_x - 0.777a_y + 0.813a_z) nN.

ii) Electric field E = F / Q. Since Q = 1 nC, E = (-0.606a_x - 0.777a_y + 0.813a_z) V/m.