Question

Three identical metal spheres are hung from a ceiling on rigid non-conducting rods. Sphere 1, sphere 2, and sphere 3 have charges 71 µC, −52 µC, and 46 µC, respectively. Each sphere is separated by 17 cm.
(a) What is the net electric force on the middle sphere due to spheres 1 and 3? (Enter the magnitude only.)(b) What is the direction of the net electric force?
a. up
b. down
c. left
d. right
e. no direction (zero magnitude)

Answer 1

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:

$F_{e}=k_{e}\frac{q_{1}q_{2} }{r^{2} }$

where $k_{e}$=$8.99x10^{9}N\frac{m^{2} }{C^{2} }$

So knowing this, we can now calculate each force. Let's start with the force exerted by sphere 1 over sphere 2:

$F_{12}=8.99x10^{9}N\frac{m^{2} }{C^{2} }\frac{(71x10^{-6}C)(52x10^{-6}C) }{(0.17m)^{2} }$

Which gives me a force of:

$F_{12}=-1148.48N$

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:

$F_{23}=8.99x10^{9}N\frac{m^{2} }{C^{2} }\frac{(46x10^{-6}C)(52x10^{-6}C) }{(0.17m)^{2} }$

Which gives me a force of:

$F_{23}=744.09N$

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.

$F_{net}=$∑F

$F_{net}=F_{12}+F_{23}$

$F_{net}=-1148.48N+744.09N$

$F_{net}=-404.39N$

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.

Answer 2

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;