Problem related to Coloumb's force

[DrClaude]

I thought charges were at the ends of string.

They are, but the strings are moving.

 

[gracy]

The string is 3 meters long. Start out with the charges 1 meter apart

I thought So ,they are not at the ends of string.

 

[DrClaude]

I thought So ,they are not at the ends of string.

Yes they are, as in the picture.

 

[Doc Al]

I thought So ,they are not at the ends of string.

Sure they are.

 

[gracy]

Are we not confusing with the problem in op and the example by @Doc Al

 

[gracy]

At least I am !

 

[DrClaude]

Are we not confusing with the problem in op and the example by @Doc Al

No, he was just rephrasing the same problem, with number values for the length of the strings.

 

[gracy]

No, he was just rephrasing the same problem, with number values for the length of the strings.

How?op mentions two strings whereas @Doc Al took one string in his example

 

[DrClaude]

How?op mentions two strings whereas @Doc Al took one string in his example

Sorry, I misread.

But that string could look like the two in the OP.

 

[gracy]

could you Please guide me further.

 

[Orodruin]

could you Please guide me further.

You have already been given a significant amount of guidance and several helpers have asked you questions which would guide you in the right direction if you tried to answer them. Instead, you are ignoring these questions thinking that they cannot possibly have anything to do with your problem. If you are going to have any chance of resolving this issue, you need to start listening to what people are telling you.

Take your image as an example, can you draw the free body diagram for one of the masses? What forces are going to be acting on the mass? Which are their directions? What is the necessary requirement for equilibrium to occur?

 

[gracy]

You have already been given a significant amount of guidance and several helpers have asked you questions which would guide you in the right direction if you tried to answer them. Instead, you are ignoring these questions thinking that they cannot possibly have anything to do with your problem. If you are going to have any chance of resolving this issue, you need to start listening to what people are telling you.

I am trying my best.You would have definitely agreed if you were to see my rough /practice book.What else I can do.

 

[gracy]

I don't know whether it is correct but i think tension force will be able to be split into one horizontal force and one vertical.The horizontal one would cancel the repulsive force.But one problem with this is then vertical force would be unbalanced so net force would be vertical so motion should be in vertical direction.

l (lower case L)cos theta =kq^2/d^2
here d=distance between the bodies when horizontal component of tension cancels the repulsive force.And k is coulomb constant having value of 9 multiplied by 10^9.

as if bodies were attracting each other

It has been a long thread.What's the use?How many helpers are there,is not something that matters.A student needs approval ,corrections.

 

[DrClaude]

I'll give it one more try.

This suggests r=2l
I want to know how to find "r"how r came out to be 2l?

What does the system of charges and strings look like if $r=2l$?

 

[gracy]

I thought charges were at the ends of string.

Sure they are.

But that string could look like the two in the OP.

What does the system of charges and strings look like if r=2l?

The charges would be like the charges at the ends of string of length 2l

 

[DrClaude]

The charges would be like the charges at the ends of string of length 2l

So how do you go from two charges attached to strings of length $l$ to the two charges being $2l$ apart?

 

[gracy]

So how do you go from two charges attached to strings of length ll to the two charges being 2l2l apart?

Angle will increase until it is 180 degrees?

 

[DrClaude]

Angle will increase until it is 180 degrees?

Yes!

Can you see now why this is happening?

 

[gracy]

Can you see now why this is happening?

Because of repulsion between like charges

 

[DrClaude]

Because of repulsion between like charges

Don't forget the tension in the string.

Are you satisfied that you understand the answer to your question "I want to know how to find "r"how r came out to be 2l?"

 

[gracy]

Can you see now why this is happening?

I don't know whether it is correct but i think tension force will be able to be split into one horizontal force and one vertical.The horizontal one would cancel the repulsive force.But one problem with this is then vertical force would be unbalanced so net force would be vertical so motion should be in vertical direction.

This vertical force is responsible for going from two charges attached to strings of length l to the two charges being 2l apart
And after that tension in string would be equal to the repulsive force which has the value of (1/4πε0)*q^2/4l^2
r being 2l
Right @DrClaude ?

 

[gracy]

Are you satisfied that you understand the answer to your question "I want to know how to find "r"how r came out to be 2l?"

Seeking your "yes" in answer to my post #56 to be satisfied.

 

[DrClaude]

This vertical force is responsible for going from two charges attached to strings of length l to the two charges being 2l apart
And after that tension in string would be equal to the repulsive force which has the value of (1/4πε0)*q^2/4l^2
r being 2l
Right @DrClaude ?

Yes. When the two strings become horizontal, there is no more vertical component of the force, and equilibrium is reached between repulsion and tension.

Seeking your "yes" in answer to my post #56 to be satisfied.

We not all in the same time zone, and I need to sleep from time to time