Question

Can you help me to solve these two problems?

1. Find the force on the load Q =-3C that is located at point (6.3). Solve the exercise using MATLAB. Check the result by solving the exercise by hand.

Answer 1

Force on a Charge is given by Coulumbe U Law E = I Q, Qz 4rto î force on Q2 due to Q, → F Q (nigen (12192) v(x2-uydt cy2-yi? r = unit vector to represent force direction. (12-21) i + Cyz-y, f i V (22-2² + cys-y,)². dores Let Qi- the charge Qi= the charge Q = tive marge 22=-ve charge forces repulsive or 42320, Attractive Forces.

Electie Fredd forre per unit charge. att Morge (miny) (M2142) È I Qi ft al Unto r2 unto (89 â r- n-ni] f (ya-y,,? Ânz-20(42-y) í - Venz- nya-4,8 t +20 (194) tfach *36,3). .Q --30 (10) Actual convention followed дис Since, & is a negative charge, is a negative charge

Unlike Charges Repel force clirection should be attractive. Actual O force direction Diagram 993, & +2C (29,41 ) Club Q-3c (uz 193? +4c (ny42) But we will be solving problem per convention. per conventron as (11) . 8,- V (23.-2) + (93-91) =ro-i)²+(3-4)²

в2 = y(6 - + (3-1) , 5 + 1 24 I he value of he 4х4 1. The value of t Гар (1) 1) Чле Члке — итъ - - стоях (-3)x ( 2 ) (26 . .. 26. ах Б ач в ло". (-5 - х10 26 ) , 2011 xio" (-5

I Foil = 2.077 0 N Direction - T-sitj 526 Focuz I (928) unto urto 82² a axioa X (-3)*(4) (setzj. (52q2 rza - 5 - zi 1 - 2 = 3.72 X109 mx la I Fad 1= 3.720109 Direction, a l- 11 - via - Face & Faci Metforce of 6 =

+j For 201710" (3) ( o =2.0778109 +3.72x100(-56-23 rza Fa = 1095-210367 ? - 3.454 î+ 0.4 0739 2 - 1.3816 P ] sest Fa ²10 g (= 5.4907 -0.9743 ~ 5.5765810N 1-0.98416E-0.17 IF 5.57651109 576 de directiona -0.98 46 -0.17u7j

MATLAB CODE

clear all
clc
% Question 1
q1=2;
q2=4;
q=-3;
x=[1 1 6]; % specify x coordiantes of all three charges Q1 Q2 Q respectively
y=[4 1 3]; %specify y coordiantes of all three charges Q1 Q2 Q respectively
r1=sqrt((x(3)-x(1))^2 + (y(3)-y(1))^2); %define r1
r2=sqrt((x(3)-x(2))^2 + (y(3)-y(2))^2); %define r2
r1_cap= [(x(3)-x(1))/r1 (y(3)-y(1))/r1]; %define r1_cap in terms of i and j
r2_cap= [(x(3)-x(2))/r2 (y(3)-y(2))/r2];%define r2_cap in terms of i and j
Fqq1=(9*10^(9)*q*q1/r1^2).*r1_cap; %Calculate Fqq1
Fqq2=(9*10^(9)*q*q2/r2^2).*r2_cap; %Calculate Fqq2
Fq=Fqq1+Fqq2; %Calculate net force
display(' Net force in i and j direction respectively')
Fq
display(' Magnitude of net Force on Q =-3C')
z=sqrt(Fq(1)^2 + Fq(2)^2)
display('Direction in i and j respectively')
Fq./abs(z)

Command Window Net force in i and j direction respectively Fa = 1.0e+09+ -5.4944 -0.9758 Magnitude of net Force on Q =-3C 5.5803e+09 Direction in i and j respectively ans = -0.9846 -0.1749

m2 Q POS, +29 Ear SE Q tuc o 8,= √ (5-1) ²7 (1-4)2 =5 regon-shop 82 = √(5-18 + (1-D? = 4 unto = 4x109x2 coisa -0.6ſ) 25 25 = 0.72x109Ncoidê-0.69)

> E = 9x1094 42 É = 2.25@x10 9 Na î era eta =109(2.25 +0.7260.81-0.6;)) 2.25 + O . z vos Floor E = (2.326 @ -0.43201) TEP 1: 2.8588x 10" Nila Derection = 0.98 85 €

MATLAB CODE

clear all
clc
% Question 2
q1=2;
q2=4;
x=[1 1 5]; % specify x coordiantes of all 2 charges and point Q1 Q2 P respectively
y=[4 1 1]; %specify y coordiantes of all 2 charges & point Q1 Q2 P respectively
r1=sqrt((x(3)-x(1))^2 + (y(3)-y(1))^2); %define r1
r2=sqrt((x(3)-x(2))^2 + (y(3)-y(2))^2); %define r2
r1_cap= [(x(3)-x(1))/r1 (y(3)-y(1))/r1]; %define r1_cap in terms of i and j
r2_cap= [(x(3)-x(2))/r2 (y(3)-y(2))/r2];%define r2_cap in terms of i and j
E1=(9*10^(9)*q1/r1^2).*r1_cap; %Calculate E1
E2=(9*10^(9)*q2/r2^2).*r2_cap; %Calculate E2
E=E1+E2; %Calculate net field
display(' Net Field in i and j direction respectively')
E
display(' Magnitude of net Field at P')
z=sqrt(E(1)^2 + E(2)^2)
display('Direction in i and j respectively ')
E./abs(z)

Command Window Net Field in i and j direction respectively 1.0e+09 * 2.8260 -0.4320 Magnitude of net Field at P 2.8588e+09 Direction in i and i respectively ans = 0.9885 -0.1511