Reference no: EM131010618
Figure E2.4(a) shows the electric field at a point due to a charge which is a vector E. The magnitude of E is given by Coulomb's law E = (q/4Πε0r2) J, where q is magnitude of the charge, r is the distance between the charge and the point, and so is the permittivity constant (8.8542 x 10-12 C2/Nm2). The electric field E at any point is obtained by superposition of the electric field of each charge. An electric dipole with q = 12 x 10-19 C is created as shown in Fig. E2.4 (b).
Write a MATLAB program to determine and plot the magnitude of the electric field along the x-axis from x = - 8 cm to x= 8 cm.
Electric field at any point (x, 0) along the x-axis is obtained by adding the electric field vectors due to each of the charges. E = E+ E. The magnitude of the electric field is the length of the vector E. The problem is solved by following steps:
1. Create a vector x for points along the x-axis.
2. Calculate the distance from each charge to points on x-axis according to the equations
rms = √(0.03 -x)2 +0.032 and rps = √(0.03+ x)2 +0.032
3. Write unit vectors in the directions from each charge to the points on the x-axis as emuv = [(0.03 -x)/rms, - 0.03/rms and epuv = [(x+ 0.03)/rps, 0.02/rps]
4 Calculate the magnitude of electric field due to positive and negative charges according to Coulomb's law:
E= emmag= (q/4Πε0rms2) and E = (q/4Πε0rps2).
5. Calculate em and ep by multiplying the unit vectors by emmag and epmag.
6 Calculate E as e = em + ep,
7. Find the magnitude of e.
8. Plot e as a function of x.
Verify your answer with Working Model 2D solution:
Hint: Goto World->Force Field-> (Dropdown Electrostatics) Pair-Wise->Fx
Your answer should look like: Check the Units if the values do not match
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