Respuesta :
Answer:
The magnitude of the force on positive charges will be [tex]\bf{227.06~N}[/tex] and the magnitude of the force on the negative charge is [tex]\bf{302.7~N}[/tex].
Explanation:
Given:
The value of the charges, [tex]q = 3.5~\mu C[/tex].
The length of each side of the triangle, [tex]l = 2.9~cm[/tex].
Consider a equilateral triangle [tex]\bigtriangleup ABC[/tex], as shown in the figure. Let two point charges of magnitude [tex]q[/tex] are situated at points [tex]A[/tex] and [tex]B[/tex] and another point charge [tex]-q[/tex] is situated at point [tex]C[/tex].
The value of the force on the charge at point [tex]A[/tex] due to charge at point [tex]C[/tex] is given by
[tex]F_{CA} = \dfrac{kq^{2}}{l^{2}},~~along~CA[/tex]
The value of the force on the charge at point [tex]A[/tex] due to charge at point [tex]B[/tex] is given by
[tex]F_{BA} = \dfrac{kq^{2}}{l^{2}},~~along~BA[/tex]
The net resultant force on the charge at point [tex]A[/tex] is given by
[tex]~~~~F_{A} = \sqrt{F_{BA}^{2} + F_{CA}^{2} + 2F_{BA}F_{CA}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{A}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(1)[/tex]
The value of the force on the charge at point [tex]B[/tex] due to charge at point [tex]C[/tex] is given by
[tex]F_{CB} = \dfrac{kq^{2}}{l^{2}},~~along~CB[/tex]
The value of the force on the charge at point [tex]B[/tex] due to charge at point [tex]A[/tex] is given by
[tex]F_{AB} = \dfrac{kq^{2}}{l^{2}},~~along~AB[/tex]
The net resultant force on the charge at point [tex]B[/tex] is given by
[tex]~~~~F_{B} = \sqrt{F_{AB}^{2} + F_{CB}^{2} + 2F_{AB}F_{CB}\cos 60^{0}}\\~~~~~= \dfrac{kq^{2}}{l^{2}}\sqrt{2(1 + \cos 60^{0})}\\or, F_{B}= \dfrac{\sqrt{3}kq^{2}}{l^{2}}~~~~~~~~~~~~~~~~~~~~(2)[/tex]
The value of the force on the charge at point [tex]C[/tex] due to charge at point [tex]A[/tex] is given by
[tex]F_{AC} = \dfrac{kq^{2}}{l^{2}},~~along~AC[/tex]
The value of the force on the charge at point [tex]C[/tex] due to charge at point [tex]B[/tex] is given by
[tex]F_{BC} = \dfrac{kq^{2}}{l^{2}},~~along~BC[/tex]
The net resultant force on the charge at point [tex]C[/tex] is given by
[tex]F_{C} = 2F_{BC} \sin 60^{0}~~along~the~line~perpendicular~to~AB\\~~~~~= \dfrac{2kq^{2}}{l^{2}}\sin 60^{0}~~~~~~~~~~~~~~~~~~~~(3)[/tex]
Substitute [tex]3.5~\mu C[/tex] for [tex]q[/tex] , [tex]0.029~m[/tex] for [tex]l[/tex] and [tex]9 \times 10^{9}~Nm^{2}C^{-2}[/tex] for [tex]k[/tex] in equation (1), we have
[tex]F_{A} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N[/tex]
Substitute [tex]3.5~\mu C[/tex] for [tex]q[/tex] , [tex]0.029~m[/tex] for [tex]l[/tex] and [tex]9 \times 10^{9}~Nm^{2}C^{-2}[/tex] for [tex]k[/tex] in equation (2), we have
[tex]F_{B} = \dfrac{\sqrt{3}(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}}\\~~~~~= 227.06~N[/tex]
Substitute [tex]3.5~\mu C[/tex] for [tex]q[/tex] , [tex]0.029~m[/tex] for [tex]l[/tex] and [tex]9 \times 10^{9}~Nm^{2}C^{-2}[/tex] for [tex]k[/tex] in equation (3), we have
[tex]F_{C} = \dfrac{2(9 \times 10^{9}~Nm^{2}C^{-2})(3.5 \times 10^{-6}~C)^{2}}{(0.029~m)^{2}} \sin 60^{0}\\~~~~~= 302.7~N[/tex]
