Respuesta :
Answer:
The net force on X is Fx=75.4N
The net force on Y is FY=25.17N
Explanation:
This is an electrostatic problem, we can calculate de force applying the formula:
[tex]F=k*\frac{Q*Q'}{r^2}\\where:\\k=coulomb constant\\r=distance\\Q=charge[/tex]
the force because of charge at 1cm on the X axis, will only have an X component of force, so:
[tex]Fx1=8.98755*10^9*\frac{(9\µC)(7\µC)}{(5*10^{-2}m)^2}\\Fx1=226.48N[/tex]
For the force because of the charge of the Y axis we have to find the distances usign pitagoras, and the angle:
[tex]r=\sqrt{(-1*10^{-2}m)^2+(6*10^{-2}m)^2} \\r=6.08cm=0.0608m[/tex]
we can find the angle with:
[tex]\alpha = arctg(\frac{1cm}{6cm})=9.46^o[/tex]
We now can calculate the force of the X axis because of the second charge:
[tex]Fx2=8.98755*10^9*\frac{(-9\µC)(7\µC)}{(6.08*10^{-2}m)^2}*cos(9.46 )\\Fx2=-151.08N[/tex]
and for the Force on Y axis:
[tex]Fy2=8.98755*10^9*\frac{(-9\µC)(7\µC)}{(6.08*10^{-2}m)^2}*sin(9.46 )\\Fy2=-25.17N[/tex]
The net force on X axis is:
Fx=226.48N-151.08N=75.4N
Fy=-25.17N
Answer:
The magnitude of the resultant force is equal to 0.0216 N
The direction is along the negative y axis
Explanation:
According to the exercise data:
q1 = 9x10^-6 C (0,1)
q2 = -9x10^-6 C (0,-1)
q = 7x10^6 C (6,0)
the distance between load q1 and load q will be equal to:
[tex]r_{1} = \sqrt{(0-1)^{2}+(6-0)^{2} }= 6.08 m[/tex]
The same way to calculate the distance between q2 and q:
[tex]r_{2} =\sqrt{(0-(-1))^{2}+(6-0)^{2} } =6.08 m[/tex]
The force on q due to the load q1, is calculated with the following equation:
[tex]F_{1}=\frac{K*q_{1}*q }{r1^{2} } *(cos45i-sin45j)=\frac{8.987559x10^{9}*9x10^{-6} *7x10^{-6} }{6.08^{2} }*(cos45i-sin45j)=0.0153*(0.707i-0.707j)=0.0108N(i-j)[/tex]
The same way to force F2:
F2 = (8.987559x10^9*-9x10^-6*7x10^-6))/(6.08^2)*(cos45i-sin45j) = 0.0108 N(i-j)
The resultant force:
F = F1 + F2 = 0.0108 N *(-2j) = - 0.0216 N j
The magnitude is equal to 0.0216 N
The direction is along the negative y axis
