Answer:
The force on X Fx=0 N
The force on Y Fy=-2.18 N
Explanation:
We have an array of charges, we will use the coulomb's formula to solve this:
[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]
but we first have to find the distance and the angule of the charge respect the charges on the Y axis:
[tex]r=\sqrt{(7*10^{-2}m)^2+ (3*10^{-2}m)^2} \\r=7.62cm=0.0762m[/tex]
we can notice that it is the same distance from both charges on Y axis.
we can find the angle with:
[tex]\alpha = arctg(\frac{7cm}{3cm})=66.80^o[/tex]
for the charge of 5µC [tex]\alpha =-66.80^o[/tex]
for the charge of -5µC [tex]\alpha =66.80^o[/tex]
the net force on the X axis will be:
[tex]F_{x5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(-66.80)\\F_{x5u}=0.465N[/tex]
and
[tex]F_{x(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*cos(66.80)\\F_{x(-5u)}=-0.465N[/tex]
So the net force on X will be Zero.
for the force on Y we have:
[tex]F_{y5u}=9*10^9*\frac{5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(-66.80)\\F_{y5u}=-1.09N[/tex]
and
[tex]F_{y(-5u)}=9*10^9*\frac{-5*10^{-6}*2*10^{-6}}{0.0762^2}*sin(66.80)\\F_{y(-5u)}=-1.09N[/tex]
Fy=[tex]F_{y5u}+F_{y(-5u)}[/tex]
So the net force on Y is Fy=-2.18N
