Respuesta :
Answer:
The magnitude of this force is [tex]1.866\times10^{-4}\ N[/tex]
The direction of this force is 55.69°.
Explanation:
Given that,
Charge at origin [tex]q_{1}= -3.40\ nC[/tex]
Charge at y axis [tex]q_{2}= 2.45\ nC[/tex]
Distance on y axis = 4.25 cm
Third charge [tex]q_{3}= 5.00\ nC[/tex]
Distance on x axis = 2.90 cm
(a). We need to calculate the force F₁₃
Using formula of force
[tex]F_{13}=\dfrac{kq_{1}q_{3}}{r^2}[/tex]
Put the value into the formula
[tex]F_{13}=\dfrac{9\times10^{9}\times(-3.40\times10^{-9})\times5.00\times10^{-9}}{(2.90\times10^{-2})^2}[/tex]
[tex]F_{13}=-0.00018192\ N[/tex]
[tex]F_{13}=-1.82\times10^{-4}\ N[/tex]
We need to calculate the force F₁₂
Using formula of force
[tex]F_{12}=\dfrac{kq_{1}q_{2}}{r^2}[/tex]
Put the value into the formula
[tex]F_{12}=\dfrac{9\times10^{9}\times(-3.40\times10^{-9})\times2.45\times10^{-9}}{(4.25\times10^{-2})^2}[/tex]
[tex]F_{12}=-0.00004150\ N[/tex]
[tex]F_{12}=-4.15\times10^{-5}\ N[/tex]
The magnitude of this force is
The total force exerted on this charge by the other two charges.
[tex]F=\sqrt{F_{13}^2+F_{12}^2+2F_{13}F_{12}\cos \theta}[/tex]
Here, [tex]\theta=90[/tex]
[tex]F=\sqrt{F_{13}^2+F_{12}^2}[/tex]
Put the value into the formula
[tex]F=\sqrt{(-1.82\times10^{-4})^2+(-4.15\times10^{-5})^2}[/tex]
[tex]F=0.0001866\ N[/tex]
[tex]F=1.866\times10^{-4}\ N[/tex]
(c). We need to calculate the direction of this force
Using formula of direction
[tex]\tan\theta=\dfrac{y}{x}[/tex]
Put the value into the formula
[tex]\theta=\tan^{-1}(\dfrac{4.25}{2.90})[/tex]
[tex]\theta=55.69^{\circ}[/tex]
Hence, The magnitude of this force is [tex]1.866\times10^{-4}\ N[/tex]
The direction of this force is 55.69°.
