Respuesta :
Answer:
The net force exerted by these two charges on a third charge is [tex]5.468\times10^{-6}\ N[/tex]
Explanation:
Given that,
Third charge [tex]q_{3}=49.0\ nC[/tex]
Distance[tex]x_{3}=-1.085\ m[/tex]
Suppose The magnitude of the force F between two particles with charges Q and Q' separated by a distance d. Consider two point charges located on the x axis one charge, q₁ = -12.5 nC , is located at x₁ = -1.650 m, the second charge, q₂ = 31.5 nC , is at the origin.
We need to calculate the total force will be the vector sum of two forces
Using Coulomb's law,
[tex]F_{13}=\dfrac{kq_{1}q_{3}}{(x_{1}-x_{3})^2}[/tex]
Put the value into the formula
[tex]F_{13}=\dfrac{9\times10^{9}\times(-12.5\times10^{-9})\times49\times10^{-9}}{(-1.650-(-1.085))^2}[/tex]
[tex]F_{13}=-17268.3\times10^{-9}\ N[/tex]
We need to calculate the force will be to the negative charge with opposite charges
Using Coulomb's law,
[tex]F_{23}=\dfrac{kq_{2}q_{3}}{(x_{2}-x_{3})^2}[/tex]
Put the value into the formula
[tex]F_{23}=\dfrac{9\times10^{9}\times(31.5\times10^{-9})\times49\times10^{-9}}{(-1.085)^2}[/tex]
[tex]F_{23}=11800.2\times10^{-9}\ N[/tex]
The force also will be to the negative side, charges with same charge sign
We need to calculate the net force exerted by these two charges on a third charge
Using formula of net force
[tex]F_{net}=F_{13}+F_{23}[/tex]
[tex]F_{net}=-17268.3\times10^{-9}+11800.2\times10^{-9}[/tex]
[tex]F_{net}=-0.0000054681\ N[/tex]
[tex]F_{net}=-5.468\times10^{-6}\ N[/tex]
Negative sign shows the negative direction.
Hence, The net force exerted by these two charges on a third charge is [tex]5.468\times10^{-6}\ N[/tex]
