Question

Four charges,
q1 = +150 µC, q2 = +47 µC, q3 = −134 µC, and q4 = +22 µC,
are fixed at the corners of a 4 m by 5 m rectangle, as illustrated in the figure below. What are the magnitude (in N) and the direction (in degrees counterclockwise from the +x-axis) of the net force acting on q1? (Assume the x-axis extends from q1 to the right.)

Answer 1

Force betwwn two particales $q_1\ and\ q_2\ is\ given\ by:\\$

$F=\frac{kq_1q_2}{r^2}$

Let $q_1\ be\ A,\ q_2\ be \ B,\ q_3\ be\ C\ and\ q_4\ be\ D$

Force on A due to B :

$F_{AB}=\frac{9\times10^9\times150\times47\times10^{-12}}{4^2}=3.97\ N\ along\ -x\ axis\\Similarly,\\$

Force on Adue to C :

$F_{AC}=4.4412\ N$

Direction of this force is : $\theta_1=tan^{-1}\frac{5}{4}=$ $51.3\degree\ clockwise\ of\ x\ axis$

Similarly,

Force on A due to D :

$F_{AD}=1.188\ N\ along\ y\ axis$

resultant of $F_{AB}\ and\ F_{AD}\ \ is:$

$F_{ABD}=\sqrt{3.97^2+1.188^2}=4.144\ N\ \\Direction\ of\ this\ resultant\ is:$

$\theta_2=tan^{-1}\frac{3.97}{1.188}=73.34\degree\ anticlockwise\ of\ y\ axis$

Now net force on A is given by:

$F_{net}=\sqrt{4.144^2+4.4412^2+2\times4.144\times4.4412\cos214.64\degree}=2.57\ N$

Direction of this net force is given by ;

$\theta=121.33\degree\ clockwise\ of\ F_{ABD}\\Or,\\((90+73.34)-121.33)=42.01\degree\ anticlockwise\ of\ x\ axis$