Phys 121A – 09 Physics Laboratory Report

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Physics

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Jan 9, 2024

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Phys 121A – 09 Physics Laboratory Report #1 Lab 201: Electric Field by Point charge Kevin Butryn Rosa Roger, Sachiv Gupta 9/19/14 Zhaoqian Su 1. Code (see attached Pages) 2. Graph (see attached pages) 3. Discussion/ Questions 1. Why is it important to exclude the point x= 1 and x = -1? -The point charges are located at these locations. When the charge crosses over these points the electric field is 0 because the point charge is ontop of the charges. 2. We already know that the electric field varies as the square of the distance between two point charges. Why do the equations for electric field above vary as distance cubed in the denominator? - Because you multiply the distance in the numerator, thus the cubed distance in the denominator gets canceled out and returns to squared. 3. Make a plot of the total electric field form x = -2.1 to x = 2.1 a) Does the electric field = 0 at origin -yes b) Why would you expect it to?

-Because they are of equal charge and they are equidistant from each other. Thus the forces cancel each other out. c)Assume that the two charges remain fixed. IF you place a positive test charge at the origin (motion constrained to x axis) would the test charge be at a point of stable equilibrium? -Yes d)Why? -Because electric field is 0 at that point and no outside forces are acting on it. Thus there are no forces to make it move. 4. Change the charge q2 to q2=-1x10^-9C(negative charge and redraw the plot.) a)Does the electric filed at the origin = 0? -No b) Why would you expect it to? -You wouldn’t expect it to because one test charge is repelling and the other attracting, both in the same direction. Thus the forces are uneven thus making the charge move. c) If you place a positive test charge at the origin (motion constrained to x axis) would the test charge be at a point of stable equilibrium? -No d)Why? 4. - Because one test charge is repelling and the other attracting, both in the same direction. Thus all the forces are pushing in one directions making the point charge move.

Eps0 = 8.854*10^-12; k=1/(4*pi*Eps0); >> q1 = 1*10^-9; q2 = 1*10^-9; a=1; >> x=[-2.1:0.2:2.1]; >> E1 = q1*k*(x+a)./abs(x+a).^3; >> E2 = q1*k*(x-a)./abs(x-a).^3; >> Etotal=E1+E2 Etotal = Columns 1 through 11 -8.3631 -12.1647 -19.5752 -37.3890 -101.5628 -900.8123 896.2846 96.7539 31.9564 13.0241 3.6681 Columns 12 through 22 -3.6681 -13.0241 -31.9564 -96.7539 -896.2846 900.8123 101.5628 37.3890 19.5752 12.1647 8.3631 >> plot(x,Etotal) >> xlabel 'x'; y label 'Etotal' Undefined function 'y' for input arguments of type 'char'. >> xlabel 'x'; ylabel 'Etotal' >> grid>> title 'Total Electric field vs. X'

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q1=1e-9; q2 = 1e-9; >> esp0=8.85e-12; k=1/(4*pi*esp0); >> a=1; >> [x,y] = meshgrid(-2.1:0.2:2.1,-2.1:0.2:2.1); >> r1S=(x).^2+(y-a).^2; r1=sqrt(r1S); >> r2S=(x).^2+(y+a).^2; r2=sqrt(r2S); >> e1x=(x)./r1; e1y=(y-a)./r1; >> e2x=(x)./r2; e2y=(y+a)./r1; >> E1m=(q1*k)./r1S; >> E2m=(q2*k)./r2S; >> E1x = E1m.*e1x; E1y = E1m.*e1y; >> E2x = E2m.*e2x; E2y = E2m.*e2y; >> Ex = E1x+E2x; Ey=E1y+E2y; >> Emag=sqrt(Ex.^2+Ey.^2); >> quiver(x,y, Ex, Ey, 1) >> xlabel('x'); ylabel('y');title('Electric Field Lines and Vectors') >> hold on >> [X, Y] = meshgrid(-3:0.4:3, -3:0.4:3); >> streamline(x,y,Ex,Ey,X,Y)

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