Respuesta :
Answer:
a) The net electric force on the middle sphere due to spheres 1 and 3 is 404N. b) The direction of the net electric force is to the left.
Explanation:
In order to solve this problem, we must first draw our free body diagram (See attached picture).
By definition, two opposite charges attract each other, so the force due to sphere 1 will go to the left (being the left direction negative) and the force due to sphere 3 will go to the right (being the right direction positive).
Knowing this we can use the electric force formula to calculate each of the forces:
[tex]F_{e}=k_{e}\frac{q_{1}q_{2} }{r^{2} }[/tex]
where [tex]k_{e}[/tex]=[tex]8.99x10^{9}N\frac{m^{2} }{C^{2} }[/tex]
So knowing this, we can now calculate each force. Let's start with the force exerted by sphere 1 over sphere 2:
[tex]F_{12}=8.99x10^{9}N\frac{m^{2} }{C^{2} }\frac{(71x10^{-6}C)(52x10^{-6}C) }{(0.17m)^{2} }[/tex]
Which gives me a force of:
[tex]F_{12}=-1148.48N[/tex]
In this case the force will be negative because it's directed towards sphere 1, this is to the left.
We can do the same with the force due to sphere 3:
[tex]F_{23}=8.99x10^{9}N\frac{m^{2} }{C^{2} }\frac{(46x10^{-6}C)(52x10^{-6}C) }{(0.17m)^{2} }[/tex]
Which gives me a force of:
[tex]F_{23}=744.09N[/tex]
In this case the force is positive because it's directed towards sphere 3, this is it goes to the right.
With these two values I can now find the net force electric force on the middle sphere due to spheres 1 and 3.
[tex]F_{net}=[/tex]∑F
[tex]F_{net}=F_{12}+F_{23}[/tex]
[tex]F_{net}=-1148.48N+744.09N[/tex]
[tex]F_{net}=-404.39N[/tex]
a) Since part a of the problem only asks us for the magnitude, then the net electric force on the middle sphere due to spheres 1 and 3 is 404.39N
b) since the answer when solving this problem was negative, this means that the force will be directed towards sphere 1, this is to the left.
A) The net electric force on the middle sphere due to spheres 1 and 3 in magnitude only is; 404.85 N
B) The direction of the net electric force is;
C; Left
Force between charges
The formula for electric force between two charges is;
F = k*q1*q2/r²
Where;
k is coulombs constant = 9 × 10^(9) N.C/m²
q1 is first charge
q2 is second charge
r is distance of separation of charges.
W are given charges of the spheres as;
Charge on sphere 1; q1 = 71 µC = 71 × 10^(-6) C
Charge on sphere 2; q2 = −52 µC = -52 × 10^(-6)C
Charge on sphere 3; q3 = 46 µC
r = 17 cm = 0.17 m
Thus;
A) Force of the first sphere on second sphere is;
F_1,2 = (9 × 10^(9) × 71 × 10^(-6) × -52 × 10^(-6))/0.17²
F_1,2 = -1149.76 N
Force on second sphere due to the third sphere is;
F_2,3 = (9 × 10^(9) × -52 × 10^(-6) × -46 × 10^(-6))/0.17²
F_2,3 = 744.91 N
Net force is;
F_net = F_1,2 + F_2,3
F_net = -1149.76 + 744.91
F_net = −404.85
Since we want the force in magnitude only, then;
|F_net| = 404.85 N
B) Since the net force was negative, it means the direction will be in the direction of sphere 1 which is to the left.
Read more on forces between charges at;
