Charged hemisphere – numerical integration. The purpose of this MATLAB exercise is to in...

Charged hemisphere – numerical integration. The purpose of this MATLAB exercise is to introduce numerical integration as the third (often the only available) way to solve integrals, besides symbolic MATLAB integration and analytical solutions. Consider the uniformly charged hemispherical surface in Fig.1.9 and use MATLAB and numerical integration to compute the electric field intensity vector at an arbitrary point (for any z) along the z-axis. The simplest numerical integration formula is

where N denotes the number of integration segments (increments) and x_i (i = 1, 2, . . . ,N) are coordinates of centers of segments. (ME1_ 12.m on IR) H

HINT: We subdivide the hemisphere into thin rings, as depicted in Fig.1.9. The radius of a ring whose position on the hemisphere is defined by an angle θ (0 ≤ θ ≤ π/2) is a_r = a sin θ and its charge is given by

Where C_r and dl_r denote the ring circumference and width, respectively. From Fig.1.9, the local z-coordinate of the field point P with respect to the ring center is z_r = z −a cos θ and the distance of the point P from the source point P on the ring equals, by means of the cosine formula,

So, based on Eq.(1.15) and the superposition principle, the electric field intensity dE at the point P due to the charge dQ and the total field E can be expressed as

To solve the integral in θ in Eq.(1.22) by numerical integration, according to Eq.(1.19), the main for loop is as follows:

Compare the result, for different positions of the field point along the hemisphere axis (different values of the coordinate z in Fig.1.9) and different values of a and ρs, to the field expression obtained by analytical integration [with the substitution given by Eq.(1.21)], which reads

For different values of the integration increment dtheta in the above for loop, evaluate the relative error of numerical integration, relative to the exact (analytical) solution.

Reference: Equation (1.15)