Room for Physics

S.M.T

05:37

@JohnRennie Hii sir ,
I am trying to understand what is potential difference & electric potential energy.
[![enter image description here][1]][1]

[1]: https://i.sstatic.net/ut5is.jpg
This is the example I read online. So , +ve Charge Q has an electric field. There is a unit +ve charge B in the diagram. When same poles come near each other , repel. But in the example , we want to move point B to A.
Q1 : Why is it that they do not repel ?
In the example , they say that work has to be done on point B to move the charge to A.

John Rennie

@S.M.T Hi :-)

S.M.T

Q4 : Can we say point A is high potential & B is Low potential

@JohnRennie Hii sir.

John Rennie

Potential difference is the electric potential energy per unit charge.

The potential of the charge Q at a distance r from the charge is V(r) = kQ/r. Yes?

S.M.T

@JohnRennie K sir. I do not know about this formula yet.

V is potential , r is distance the unit charge has to cover from its original position to the position it needs to go to

John Rennie

Do you know that the force between two charges a distance r apart is F(r) = kQ₁Q₂/r² ?

S.M.T

05:44

Q is the amount of charge needed for it to go there

@JohnRennie Not exactly. But I do know it is similar to G*m1m2/r^2.

John Rennie

Yes, replace G by the electrostatic constant k, and the masses m₁ and m₂ by the charges Q₁ and Q₂ and you get the electrostatic force equation F(r) = kQ₁Q₂/r².

S.M.T

Yes sir.

Sir , from the text I posted above. Could you please verify my difficulties in it

John Rennie

Now replace Q₁ by the charge Q in your image, and make the other charge a unit charge Q₂ = 1, and we get the force between the charge Q and a unit charge: F(r) = kQ/r².

Yes?

@S.M.T I will answer your questions but we need a bit of background first.

S.M.T

@JohnRennie K sir.

@JohnRennie k sir

John Rennie

Suppose we move our unit charge a small distance dr closer to Q. The force is F(r) = kQ/r² and work is force times distance, so the work we need to do is:

dW = F(r) dr = F(r) = kQ/r² dr

OK so far?

S.M.T

05:56

@JohnRennie K.

@JohnRennie @JohnRennie What will be r^2 here sir ?

Is the r in the formula = dr ?

John Rennie

r is the distance between the 2 charges, and dr is the small distance we move our unit charge towards Q.

I can draw a diagram if you want.

S.M.T

@JohnRennie K sir. r is the distance we want to unit charge to move & dr is just a small distance in that r.

John Rennie

That's what I mean by r and dr

The green sphere is the unit charge we are moving towards Q.

Since the distance between the charges is r, the force between them is F(r) = kQ/r². Yes?

S.M.T

Force between them means delta F I.e kQ/r^2(For green) - kQ/r^2(red) ?

John Rennie

The force between them is just the force you would need to apply if you were holding the two charges in place.

S.M.T

06:09

Ok sir. But if we say force between them , then it means both of them apply some force right.

John Rennie

The red sphere applies a force to the green sphere, and by the third law the green sphere applies an equal and opposite force to the red sphere. Yes?

S.M.T

@JohnRennie Yes

John Rennie

And that force is F(r) = kQ/r².

S.M.T

@JohnRennie K. So , for individual values of force for green sphere & red sphere ; values of Q , r are different for each of them ? Like green will have higher r than red but red will have higher Q than green.

John Rennie

I don't understand what you are asking ...

S.M.T

06:13

@JohnRennie is it better now ?

John Rennie

In the question we start with the unit charge (the green sphere) at a distance r = B and we move it towards the charge Q until the distance has decreased to a distance r = A. Yes?

S.M.T

@JohnRennie Yes sir.

John Rennie

And what I am going to do is calculate how much work is needed to move the unit charge from B to A.

S.M.T

@JohnRennie yes

John Rennie

So what I'm saying is that if the charges are a distance r apart then the force between them is F(r) = kQ/r², so if we move the unit charge a small distance dr the work done is dW = F(r) dr = kQ/r² dr

OK so far?

S.M.T

06:19

Yes sir

John Rennie

And to find the total work needed to move from B to A we need to integrate ∫ dW. Yes?

S.M.T

@JohnRennie yes

John Rennie

If we integrate F(r) = kQ/r² we get ∫ kQ/r² = - kQ/r

S.M.T

Ok

John Rennie

I'm going to ignore the minus sign for now (I can explain why later) and just write this as kQ/r.

S.M.T

06:22

k sir.

John Rennie

Then because we are integrating from B to A we get the work needed to move from B to A is:

W = kQ/A - kQ/B

S.M.T

@JohnRennie we integrated because after every distance dr covered , charge value changes. Right

@JohnRennie K.

John Rennie

As we move the distance r is continually changing. Yes?

S.M.T

@JohnRennie Yes

John Rennie

And as r changes the force changes. That's why we need to integrate.

S.M.T

06:24

@JohnRennie KQ/B - KQ/A = KQ/r in total right ?

@JohnRennie Ok sir.

John Rennie

>KQ/B - KQ/A = KQ/r in total right ?

I don't understand what you are asking here ...

S.M.T

I got it sir. Let’s move ahead

John Rennie

OK so the work is W = kQ/A - kQ/B, but we can get this in a simpler way.

S.M.T

K

John Rennie

We define the potential at a distance r from Q as V(r) = kQ/r.

S.M.T

06:28

Yes

John Rennie

And then the work needed to move from B to A is just the difference in the potential at A and the potential at B i.e. W = V(A) - V(B).

S.M.T

K

John Rennie

Since V(r) = kQ/r we get W = kQ/A - kQ/B, and that's exactly what we got by integrating the force. Yes?

S.M.T

Yes

John Rennie

The potential and the work done are basically the same thing, which is why we got the same answer.

S.M.T

06:30

@JohnRennie K. So, this is known as potential difference

John Rennie

Yes. You started out asking:

> I am trying to understand what is potential difference & electric potential energy

S.M.T

Yes sir.

John Rennie

And the point I'm making is that they are the same thing.

S.M.T

Potential difference & work done & electric potential energy are same thing ?

John Rennie

Yes

S.M.T

06:33

@JohnRennie Few questions of mine . Q1 : Do Q1 & Q , Both have electric fields ? , Q2: Does the red charge also move ?

John Rennie

Both charges have an electric field. The force actually comes from the way the two electric fields interact, though we don't need to worry about the details of this.

S.M.T

K

John Rennie

Q2: we are assuming something is holding the red charge Q so it is fixed in place. i.e. we are moving the unit charge (green) towards Q and Q stays fixed.

S.M.T

K.

John Rennie

It gets a little more complicated if both charges move, though the basic principles remain the same.

S.M.T

06:36

Q3 : What moves them is electric field ?

The force is applied by electric field on them right

John Rennie

I am assuming we are applying some external force to move the charges together.

Suppose both charges are positive. Like charges repel, so the two positive charges will push each other away.

S.M.T

@JohnRennie Yes

John Rennie

That means if we want to move the charges closer together we have to apply an external force. The work I calculated by integrating is the work done by this external force.

S.M.T

@JohnRennie This external force is actually equal to : External force - repulsion force = The answer we get as external force.

John Rennie

We apply an external force ad that external force does work i.e. it uses energy. That energy has to go somewhere, it can't just disappear, and where it goes is into the potential energy of the two charges.

S.M.T

06:43

@JohnRennie @JohnRennie Also sir , Why did we not consider the minus sign here ?

John Rennie

That is, the work done is equal to the change in potential energy of the charges.

@S.M.T I don't want to go into that right now.

S.M.T

K sir,

K sir. Thank you very much for today. See you in some time.

John Rennie

OK :-)

S.M.T

09:55

@JohnRennie Hii sir.

How would we define EMF

John Rennie

It depends what you're asking.

There is a strict technical definition.

But also EMF tends to be used to just mean "voltage".

S.M.T

K.

John Rennie

We've seen how charges create potential differences. Yes?

S.M.T

Yes

John Rennie

But other things can create potential differences as well. For example in a battery a chemical reaction creates a potential difference between the terminals of the battery.

S.M.T

09:59

Yes

John Rennie

The term electromotive force (usually abbreviated to EMF) refers to a potential difference created by something that isn't a charge e.g. like a battery.

S.M.T

@JohnRennie K.

John Rennie

So strictly speaking when we are talking about the voltage of a battery we should say the EMF of the battery.

(Though no-one ever does :-)

S.M.T

When no current is drawn from a cell I.e open circuit. The potential difference between the terminals is called

I got 2 definitions online.

Work done per unit charge in taking a +ve charge around the complete circuit of cell

2nd one is clear.

the 1st , it says online that EMF is a special type of potential difference.

@JohnRennie It is clear now.

John Rennie

OK :-)

S.M.T

10:03

@JohnRennie Sir , what will be terminal voltage?

John Rennie

Do you know what the internal resistance means for a battery?

S.M.T

@JohnRennie Not yet sir.

@JohnRennie Sir ,

opposition to the flow of current within the battery

so , just different in definition.

nternal resistance is the rsisitance offered by the electrolyte to the flow of electrons.

John Rennie

10:19

Hi, sorry I was on the phone.

If we have a current I flowing through a resistance R then the voltage decreases by IR. Yes?

S.M.T

11:03

@JohnRennie Yes

@JohnRennie Ok.Np sir.

John Rennie

Hi

S.M.T

Hii sir.

John Rennie

OK suppose the battery has an EMF E, but it has a resistance r (it's common to use small r for the resistance of a battery).

We call this the "internal resistance" because it's the resistance inside the battery.

Then if a current I is flowing through the battery the voltage decreases by Ir due to this internal resistance r, so the voltage we measure at the terminals of the battery is V = E - Ir.

OK so far?

S.M.T

@JohnRennie E & V are same thing right ?

Only difference is that E is voltage in battery

& V is in circuit

John Rennie

V is the voltage we measure at the terminals of the battery if we connect a voltmeter to it.

V = E - Ir so if there is no current, i.e. I = 0, then we get E = V.

But if there is a current flowing we get E > V.

This voltage V is the terminal voltage.

S.M.T

11:12

V = IR , This V is terminal voltage

John Rennie

Is the R the internal resistance that I talked about above?

S.M.T

@JohnRennie Ohk. No. I’m sorry. I went wrong.

@JohnRennie K. I got till here sir.

John Rennie

You started out asking:

1 hour ago, by S.M.T

@JohnRennie Sir , what will be terminal voltage?

S.M.T

Yes

John Rennie

And what I'm saying is that the terminal voltage is V = E - Ir

where E is the EMF of the battery, I is the current flowing through the battery and r is the internal resistance of the battery.

S.M.T

11:17

K sir. @JohnRennie

I got it.

John Rennie

11:28

:-)

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