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midterm1a_solution

# midterm1a_solution - q2 =-1.0*q1%assign q2 as negative of...

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Introduction to Computational Physics Phy 265 Midterm 1a (In-Class) Solution function [potential_table]=potential(q1,d) %POTENTIAL - a function to compute the potential on the x-axis %of a dipole centered about the origin and on the x-axis. %Call syntax - [potential_table]=potential(1.60e-19,1.53e-10) %Input - magnitude of a charge in C and separation between the charges in m %Output - plot of x vs. V and a table of x (column 1) vs. V (column 2) in V %File written by S.C. Tegler. Last modified 2/21/2011 % --------------------------------------------------------------------- format short e %numerical output as short sci notation
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Unformatted text preview: q2 = -1.0*q1; %assign q2 as negative of input q1 k = 8.988e9; %Coulomb constant in MKS X1 = -d/2.0; %x location of q1 X2 = d/2.0; %x location of q2 XP = -3.0*d:0.1*d:3.0*d; %x positions for potential,VP,calculations VP = k*q1*(sqrt((XP-X1).^(2))).^(-1) + k*q2*(sqrt((XP-X2).^(2))).^(-1); plot(XP,VP, 'sk' ) %begin plotting xlabel( 'x (meters)' ); ylabel( 'V (volts)' ); title( 'Dipole Potential' ); %end plotting XP=XP' %transpose row to col vector VP=VP' %transpose row to col vector potential_table=[XP VP] %print position, potential table to %command window...
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