@user8718165 For measuring an electrode potential,I am pretty sure that some medium would be required to function as an electrolyte. Since the OP says that he had just connected the multimeter to the rods directly, I feel that cathodic protection may not be the phenomenon here

@YUSUFHASAN yes...that makes sense...then I'm not sure:-)

@ScientistSmithYT hi. I'm generally around from 05:00 to 13:00 UK time, so I've only just seen your question.

@ScientistSmithYT I've no idea what is causing the 0.6V potential difference you measured. You can get voltages developed by metal junctions, and indeed this is how thermocouples work, but 0.6V seems a very high voltage for this effect.

@user8718165 @user8718165 I guess it could be an electrochemical effect, a cell with iron and aluminium electrodes would certainly generate an EMF. But as @YUSUFHASAN says you'd normally need some form of electrolyte present.

@JohnRennie Good morning Sir

@JohnRennie yes....got it

@user8718165 morning :-)

@JohnRennie sir I want to ask you something

@user8718165 yes ... ?

@JohnRennie mixed lawn tennis has to be played among 7 couples such that no husband and wife play in the same set...

...and ... ?

@JohnRennie find all possible sets

Do all matches have to be male-female? i.e. no male-male or female-female matches?

on a particular side there has to be both a girl nd a boy...

@JohnRennie not sure...sir...nothing mentioned

@user8718165 Ah, you mean it's mixed doubles?

@JohnRennie yeah sir

So you are selecting two men and two women from seven men and seven women, subject to the criterion that the selection doesn't include a married couple.

@JohnRennie yes sir...

@JohnRennie the first person can be selected in $14$ ways...I think

I would guess you use a similar process to yesterdays question i.e. start with all possible combinations and subtract the combinations that include at least one married couple.

The number of possible combinations would be $P(7,2)^2$

@JohnRennie sir answer is 840

@JohnRennie that method doesn't work...I tried it with 2 or 3 problems...and failed

I don't know. Sorry.

How about this: choose the two men first. The number of ways of doing this is 7 x 6 = 42.

Now eliminate the two men's wives leaving five women. The number of ways of choosing the women is then 5 x 4 = 20.

@JohnRennie yes sir....I was thinking the same way

So the total number of combinations is 42 x 20 = 840.

@JohnRennie got it sir

I solved 6 more of the same kind:-) @JohnRennie

@JohnRennie Sir I have some more questions

@user8718165 yes ... ?

@JohnRennie in permutations....order doen't matter...right?

imgur.com/EvWVRuN tennis court

@user8718165 For permutations order matters. For combinations the order doesn't matter.

@JohnRennie oh yes....so sorry

@JohnRennie so in the diagram.....for 2 particular boys...the boys can stand in A-C, A-D, B-C, B-D

@user8718165 the tennis problem?

@JohnRennie yes sir

I don't think you are being asked how to arrange the people on the court. You're just being asked how to choose the four people.

7 x 6 x4

@JohnRennie oh...got it sir....I was thinking that doing that would again generate 4 separate permutations...thank you sir

@JohnRennie sir some doubts in physics

@user8718165 yes ... ?

@JohnRennie sir in the double cone problem...the rails are always tangent to the cones...right

@user8718165 yes, though note that there isn't a single tangent to a point on the cones surface. There are a whole family of tangents at different angles.

@JohnRennie okay sir...got it....so if we change the angle of the rails while keeping the contact point fixed....how does the direction of force change?

I'm offline for a bit while my PC installs a Windows update. Back soon - I hope :-)

@JohnRennie okay sir

@user8718165 I'm back. My PC appears to have updated and stayed alive :-)

@JohnRennie wow sir!a lovely PC

@JohnRennie sir chat.stackexchange.com/transcript/message/51039077#51039077

I don't think this is easy to describe. It's always hard to visualise what's going on in 3D geometry without having models in your hands that you can play with.

@JohnRennie Sir I'll try to visualize:-)

@JohnRennie whenever you've got time to answer (not interrupting you guys): we said yesterday that it's not true then that the emf source as a function of time is equal to the sum of the voltages across each of the circuit elements as a function of time. but is it true (by Kirchoff's laws) that the sum of the the sum the voltage amplitudes across the three elements equals the voltage amplitude of the power source?

@kylecampbell Instantaneous amplitude yes. i.e. suppose you have an AC voltage source $V\sin\omega t$ and a resistor and capacitor in series the it is true that $V\sin\omega t = V_R \sin\omega t + V_c \sin(\omega t + \pi/2)$.

and likewise for a simple series RLC circuit?

Yes, for an RLC circuit you'd get $V\sin\omega t = V_R \sin\omega t + V_c \sin(\omega t + \pi/2) + V_L \sin(\omega t - \pi/2)$

(I wouldn't swear under oath I have the $+\pi/2$ and $-\pi/2$ the correct way round)

ah I see

yeah, not a big deal

Vector addition of R, L and C is the instantaneous voltage, right

@Aladdin yes

So we need to know the angle between these vector for Instaneou s?

Yes

That's what the phasor diagrams tell you.

Is it generally given / visible in the question

You'll do hundreds of phasor questions while preparing for the JEE :-) They are all pretty straightforward once you get the hang of it.

@JohnRennie sir...one last PnC problem...need your help

@user8718165 hi

@JohnRennie hello sir...

@JohnRennie ALLAHABAD...how many differnt words can be formed where both L don't come together

It's P(9,9) minus the number of ways the two Ls can be together.

@JohnRennie done...thank you sir

@user8718165 cool :-)

@JohnRennie Ok thanks. I'm now on Hawaii time. But only for 10 days. So I'll reply when I can. What is an electrochemical reaction and what is going on that makes it do this?

@ScientistSmithYT the classic electrochemical reaction is the zinc copper battery. This consists of a zinc rod in a zinc chloride solution, and a copper rod in a copper sulphate solution, joined by a conducting link.

@JohnRennie I measured the voltage in two scenarios. I know that saliva has hydrochloric acid in it, but at a very low percentage. I tested it without any at first and got 0.6 volts. And then I licked it and the voltage went up a tiny bit to 0.63 volts. I see potential in this experiment of sorts. But I don't understand why, how and what does what to what that makes it work?

@JohnRennie Ok, yeah I've done that kind of battery before. But even with that I didn't have enough information that I needed to really understand how it works.

What happens is the zinc metal dissolves into solution and this reaction produces electrons. At the other side of the cell the copper ions in the solution are forced out of solution as metallic copper and this consumes electrons. So overall the reaction drives electrons from the zinc round an external circuit to the copper. This is basically how a battery works.

Oh! Ok

Is that the same for this thing I accidentally did?

The zinc copper cell is a well controlled electrochemical reaction with a precise voltage that we can measure. But any time you have different metals together you can get rather less controlled reactions of a similar sort.

Ok

In the case of zinc and iron you get a similar reaction that is the reason why zinc plating stops iron from corroding.

The zinc corrodes and generates electrons, just like in the zinc copper cell, and those electrons drive any Fe ions back out of solution as metallic iron.

The net result is that as long as any zinc metal is left the iron can't dissolve i.e. can't corrode.

@JohnRennie Sir so basically user ScientistSmithYT created a battery of some sort....is it?

@user8718165 that is certainly a possibility. What exactly is the source of the voltage is hard to say.

Oh! Ok, I'm getting it a little more

@JohnRennie that'd require a thorough inspection and knowledge of the surrounding conditions

I measured it with my multimeter set to DC. When I put it on AC it read nothing. I don't know if that means anything.

@user8718165 What do you need to know about the surroundings?

That makes sense. Cells produce a DC voltage not an AC voltage.

It definitely does not act like static electricity. I know that for certain.

@ScientistSmithYT I meant if there were some sort of electrolye of stuff....But here you don't so the exact reason is hard to say....Well I think the humid atmosphere should have some effect...but I'm not sure

I live and did this experiment in Utah. It is very very dry. Humidity is usually around 10 percent

@ScientistSmithYT have you tried hooking up the setup to a low voltage LED

@user8718165 No I haven't. Should I?

@JohnRennie Will hooking it up to an LED help sir?

@user8718165 as a general rule these kinds of accidental cell tend to have high internal resistances so they don't produce enough power to do much. In any case it would take more than 0.6V to illuminate an LED. The junction voltage is probably greater than 0.6V.

@JohnRennie My multimeter has an amperage thing on it I can use.

@JohnRennie that's a dead end:-(

@JohnRennie It does AC and DC amps

@ScientistSmithYT it might be interesting to see what current is produced. It will be a DC current so use the DC current setting. I bet it's only a few milliamps though.

@ScientistSmithYT I bet you're badly perplexed by this experiment...try asking it on Elec.Engg SE and may be on Physics SE :-) (if you wish)

@JohnRennie Ok, I'll do that after my vocation. I only have notes I took. But I'm a thousand miles away from home. So after 10 days I'll come back to you with my results.

@user8718165 I really am. I'm really curious and I've spent many sleepless nights thinking about it. But I can't think of an answer or find an answer.

@user8718165 Could you give me the link to that Elec.Engg SE site?

@ScientistSmithYT electronics.stackexchange.com/questions/ask here you have it

@user8718165 Hey thanks, I'll go ahead and ask there.

@user8718165 It poped up with this message when I went to the link. "Sorry, we are no longer accepting questions from this account. See the Help Center to learn more."

@ScientistSmithYT Sorry! I can't help more :-)

@user8718165 I read through the explanation and basically I've been banned from asking questions. And to help get the ban removed by the system I have to make my other questions better. But I can't make them better.

@user8718165 That's alright, you've definitely helped as much as you could and really, I don't expect people to go out of there way to respond. But Thank you for your time and knowledge.This also goes for @JohnRennie as well. Thank you for your time and knowledge. I'll get those results back a couple days after 10 days from now.

@ScientistSmithYT alternatively you can answer questions for removing the ban. It works too

@JohnRennie I had doubt in the ring question... I can't get the time period even after binomial expression simplification

@Aladdin hi

Hi

Yes this one

The potential is given by $V(x) = -kqQ/r$ where $Q$ is the ring charge $Q = 2\pi a \lambda$

Yes

And $r = \sqrt{a^2 + x^2}$

Yes

Shall we go with the potential as I suggested yesterday, or do you want to try using the force?

Isn't this potential energy though

I would like with force I guess

Yes, we can use either the potential or the force. For a simple harmonic oscillator $F = -kx$ and $V = \tfrac{1}{2}kx^2$ so we can get $k$ either way.

If we substitute for $r$ we get:

$$ V(x) = \frac{kqQ}{\sqrt{a^2 + x^2}} $$

Ok

And to get the force we use $F = -dV/dx$ so we have to differentiate the potential wrt $x$

Better with potential then

I get the force to be:

$$ F = -kQq \frac{x}{(a^2 + x^2)^{3/2}} $$

I got this too

So we write this as:

$$ F = kqQ x \frac{1}{a^{3/2}} \left( 1 + \frac{x^2}{a^2} \right)^{-3/2} $$

@Aladdin do you agree so far?

I get $a^3$ instead of $a^3/2$

Ah, yes, oops:

$$ F = kqQ x \frac{1}{a^3} \left( 1 + \frac{x^2}{a^2} \right)^{-3/2} $$

Yes

$$ \left( 1 + \frac{x^2}{a^2} \right)^{-3/2} \approx 1 - \frac{3}{2}\frac{x^2}{a^2} $$

If $x \ll a$

$$ F \approx kqQ x \frac{1}{a^3} \left( 1 - \frac{3}{2}\frac{x^2}{a^2} \right) $$

Yes

Ah, that's interesting. For $x \ll a$ that becomes:

$$ F = \frac{kqQ}{a^3} x $$

Hmm so we needed to ignore the x/a term

@Aladdin because if $x \ll a$ then $x^2/a^2 \ll 1$

So if we write $F = -Kx$ (using a large $K$ to distinguish it from the electrostatic constant $k$) we get:

$$ K = \frac{kqQ}{a^3} $$

OK the answer matches finally

hii @JohnRennie

@Aladdin cool :-)

@Aladdin do you want to try using the potential to check that gives the same answer?

@user8718165 hi

Yea I am on it

@JohnRennie hello sir! I'll wait....You can help user Aladdin :-)

@JohnRennie I don't get from potential method

@Aladdin start from:

$$ V = \frac{kqQ}{\sqrt{a^2 + x^2}} $$

Rearranging gives:

$$ V = \frac{kqQ}{a}\left( 1 + \frac{x^2}{a^2}\right)^{-1/2} $$

OK so far?

Yes

Using a binomial approximation we get:

$$ V = \frac{kqQ}{a}\left( 1 - \frac{x^2}{2a^2}\right) $$

Yea

$$ V = \frac{kqQ}{a} - \tfrac{1}{2} \frac{kqQ}{a^3} x^2 $$

Okay but I tried approximation here

Now the first term is just a constant. It's the potential at the centre point relative to infinity. We can subtract this off because we want our potential to be zero at the centre. That means we get:

$$ V = - \tfrac{1}{2} \frac{kqQ}{a^3} x^2 = -\tfrac{1}{2} K x^2 $$

So as before we get $K = kqQ/a^3$

We can subtract this off because we want our potential to be zero at the centre.

Is this because the potential energy at the equilibrium point is zero?

You can set the zero point of a potential energy anywhere you want, because only changes in PE matter. The absolute value of the PE is irrelevant.

For electrostatics we generally take the zero to be at infinity i.e. at infinite separation.

Wjy do we want potential zero at the center here

So in this case we write $V = -kqQ/r$ and this goes to zero as $r \to \infty$. OK so far?

Ok

With a simple harmonic oscillator we normally choose the PE to be zero at the equilibrium point. e.g. for a spring we take the PE to be zero at the unstretched length. Yes?

Yeah

But our expression $V = -kqQ/r$ would give $V = V = -kqQ/a$ at the centre of the ring i.e. at the equilibrium point. So to make the potential zero at the centre we have to add $kqQ/a$ (i.e. subtract $-kqQ/a$).

So that's what we did okay

Yes

I guess it was that subtracting off the PE at the centre that gave you the problem yesterday.

Yeah

Using the force avoids that, but then you have a differentiation to do. The differentiation isn't all that hard, but it's easy to make mistakes in an exam.

But I guess the point here to remember the basic principle i.e. approximate the motion as simple harmonic near the equilibrium point. That's quite a common approach.

I guess we could have approximated the potential first then differentiated it ...

Yeah but we can't approximate after diffrentiating.. That's a bummer

Well if you start from:

$$ V = \frac{kqQ}{a}\left( 1 - \frac{x^2}{2a^2}\right) $$

then you can differentiate to give:

$$ \frac{dV}{dx} = - \frac{kqQ}{a}\frac{x}{a^2} $$

Which ends up giving you the same result.

Ohkay. So it's same anyways

@Aladdin yes

I had some doubts in other questions too. Can I ask them now

@Aladdin yes

I had doubts in both 17 and 18...

Let's have a look at 17 ...

Tension minimum I guess electrostatic force will equal weight force

If there were no field you could calculate the centripetal force as a function of the angle $\theta$. You just calculate the velocity from the potential energy change and use $F = mv^2/r$. Yes?

Yes

Call the centripetal force $\mathbf F_c$

Ok

Now there's an electrostatic force $q\mathbf E$ that points horizontally right, so the component of this force along the string is $q E \sin\theta$.

So the total centripetal force is $F_c + qE\sin\theta$

Ok

So you need to work your expression for $F_c$, add $qE\sin\theta$ to it and find where that force is a minimum.

It's going to be on the left side of the circle isn't it, because that's where the electric field is pulling the ball inwards.

Oh wait, I haven't included the effect of the electric field on the motion of the ball.

When calculating the velocity you need to include both the gravitational and electric potential energy.

Okay

I don't see electrostatic potential energy here....

Suppose we take the PE at the bottom to be zero, and assume the velocity at that point to be $v_0$

Okay

Then the gravitational potential energy is $V_g = mgh = mg\ell(\cos\theta - 1)$

Yes

The electrostatic PE is $qEx$ where $x$ is the horizontal distance moved (right is positive $x$)

So $V_e = qE\ell\sin\theta$

I'm going to have to be careful about the signs here ...

@JohnRennie, Hello

Ahh okay

@pi-π hi, I'm in the middle of a problem with Aladdin

@Aladdin so we're going to get $V = mg\ell(\cos\theta - 1) + qE\sin\theta$

@JohnRennie, Oh..sorry.. carry on.

And that's going to be equal to the change in KE.

Okay that makes sense. Got it

I'll have to leave that with you as I have to go.

Okay. I will try to solve this

@Aladdin for Q18 just take moments about the point where the wheel contacts the slope.

How will that help

The instantaneous point of rotation of the wheel is where the $-q$ charge is. Yes?

Yes

So the weight of the wheel produces a clockwise torque and the force on the +q charge produces an anticlockwise torque. You need the two torques to be equal and opposite for the wheel not to move.

that's no slipping condition if I recall

None of the answers include the coefficient of friction, so it's implied that the friction is high enough for the wheel not to slip.

i.e. you only need consider rolling and not slipping.

Okay

@pi-π is it a quick question?

@JohnRennie

@pi-π I don't think I understand the question ...

@JohnRennie, ah. Okay

The question doesn't make sense ...

I need to go now anyway. Back tomorrow.

@JohnRennie Hi!

How's life?

:-)

@tatan hi :-)

Life is quiet right now. Everyone is on their summer holiday :-)

Are the servers also on holiday? xD

@tatan sadly not!

How's the weather in UK? It was raining a lot during the world cup :-(

In my part of the world the weather has been very good, but Chester is lucky as it doesn't get much rain. It's in the rain shadow of the Welsh mountains and the Peak District so most of the rain falls on the hills beore it reaches us.

I guess you're right in the hottest bit of the Indian summer at the moment.

yeah... with extreme rain deficit in parts and floods in other... both are intolerable... in my part its super hot,humid with almost no rain till now... in the past 10 years we never have had such a dry July :-(

if interested just see this-imdkolkata.gov.in/acwc/rfdistribution/view#n24 and this- imdkolkata.gov.in/acwc/rfdistribution/view#n23

@JohnRennie Hello sir, after a long time! :-)

@Aladdin Thanks, I'll try to answer questions to remove the ban.

@JohnRennie What is your education and experience background? I rarely find your teaching method. I really really love your teaching method.

@JohnRennie You don't have that strict rude and weird teaching method. The method where they teach and no one learns. I have only found your from of teachibg method 3 times over my 19 years of life. More people should be teaching like you do.

@JohnRennie Other people that are professors teach terribly. They teach with a prior opinion that they try to force onto there students, but you dont. It is a very rare teaching method. And I appreciate that.

@JohnRennie You taught me what takes professors days to teach me in a few minutes.

@All Does anyone know about how a high powered laser can move and spin graphite with a North South magnet configuration?