Respuesta :
Answer:
The force exerted by three charges on the fourth is [tex]F_{resultant}=2.74\times10^{-5}\ \rm N[/tex]
Explanation:
Given:
- The magnitude of three identical charges, [tex]q=10\ \rm nC[/tex]
- Length of the edge of the square a=3 cm
- Magnitude of fourth charge ,Q=3 nC
According to coulombs Law the force F between any two charge particles is given by
[tex]F=\dfrac{kQq}{r^2}[/tex]
where r is the radial distance between them.
Since the force acting on the charge particle will be in different directions so according to triangle law of vector addition
[tex]F_{resultant}=\sqrt ((\dfrac{kQq}{L^2})^2+(\dfrac{kQq}{L^2} })^2)+\dfrac{kQq}{(\sqrt{2}L)^2}\\F_{resultant}=\dfrac{kQq}{L^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=\dfrac{9\times10^9\times10\times10^{-10}\times3\times10^{-9}}{0.03^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=2.74\times 10^{-5}\ \rm N[/tex]
