Calculating Electric Forces with Dangling Objects

• Cory
In summary, the conversation is discussing how to calculate the initial charge on an identical metal sphere B when it is brought into contact with a neutral metal sphere A that is hanging from an insulating wire. The conversation delves into using equations for electric forces, force of tension, and force of gravity to solve for the initial charge on B, with the final equation being q=sqrt(Fg*tan12/r^2/k). The use of trigonometry and the distance between the two particles is also mentioned.
Cory
How do I calculate Electric forces when objects are dangling from insulated wire. For example
Neutral metal sphere A, of mass 0.10kg hangs from an insulating wire 2.0m long. An identical metal sphere B, with charge -q, is brough into contact with the sphere A. The spheres repel and stelle down as shown in the following figure

.\ <) = 12
...\
...\
...\
(B)...(A)
* <) AB(top) = 90 degrees.
Calculate the initial charge on B.

The formula's are Fe =
k*q1*q2
r^2

Ok Cory,a are you able to draw a FBD of the forces acting on A?

I did with electric force outward, Force of tension angled at 12degrees, and force of gravity. Then I rearranged them to make a triangle.
ie
^
.\ =Ft
...\
...\
...0---->Fe
...|
...| Fg
...\/

That's perfectly fine, what's your problem with solving the question then?

I can't seem to understand how to solve the FBD for fe. I tried using
tan(12) = Fe
.....m*g
but that doesn't seem to work, and I am really just not sure how to go about this.

Last edited:
That doesn't seem to be right, why would you use sin(12)=Fe?

:)

sin(12)=Fe/Ft isn't it? Think of a triangle. Thus Fe=Ft*sin(12) You can find Ft using Fg

Sorry, I edited my last post, I typed the wrong thing. Should I still try Fe = ft*sin12? I don't really understand

I don't think so.. I'm trying to find the initial charge on A

Nm I see your problem, you do know Fe.

The spheres are equal so q1 = q2

Thus:
Fe = Fg*tan12
and
Fe =k*q^2*r^2

so find Fe with mg*tan12 (which is: 0.20830543) then use that to solve for q in Fe =k*q^2*r^2? why does it change from divided by r^2 to multiplied?

Cory said:
so find Fe with mg*tan12 (which is: 0.20830543) then use that to solve for q in Fe =k*q^2*r^2? why does it change from divided by r^2 to multiplied?

I seem to still be missing radius

Heh sorry, it does not mean radius, r means the distance between the two particles. Are you able to find it using trigonometrics?

well yea, but then it's the same as Fe = Fg*tan12 which doesn't work. Maybe I should use the length of the rope?

What do you mean?

k*q^2*r^2=Fg*tan12

q=sqrt(Fg*tan12/r^2/k)

What are electric forces?

Electric forces are forces that arise between charged particles, such as protons and electrons. They can be attractive or repulsive and are caused by the interaction of electric fields.

What is Coulomb's Law?

Coulomb's Law is a fundamental law in physics that describes the force between two charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

How do electric forces affect objects?

Electric forces can affect objects by causing them to repel or attract each other, depending on the charges of the objects. They can also cause objects to move or accelerate if they are not balanced by other forces.

What is the difference between electric and gravitational forces?

The main difference between electric and gravitational forces is that electric forces act between charged particles, while gravitational forces act between objects with mass. Additionally, electric forces are much stronger than gravitational forces.

How can electric forces be used in everyday life?

Electric forces have many practical applications in everyday life, such as in the functioning of electronic devices, the generation of electricity, and the attraction and repulsion of magnets. They are also used in medical equipment, such as MRI machines, and in industrial processes, such as electroplating.

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