Problems In Electrostatics/Coulombs Law/Eequilibrium of 2 point charges/ NEET/JEE.

Added: 2026-06-02

[00:00:00]Dear students, so I'm back with the one more beautiful problem on Coulomb's law.

[00:00:07]So, the question is like this.

[00:00:09]Equilibrium of two point charges in liquid.

[00:00:12]Okay?

[00:00:13]The last time I got for you at home problem discuss we have discussed this problem last time.

[00:00:19]Take it easy that's equilibrium of two point charges.

[00:00:22]Okay?

[00:00:23]In this very problem we were supposed to calculate number one is that this angle theta and the tension of the string. And we are done with this problem, okay?

[00:00:32]Now, the question number two is connected with this. Okay, what's the problem exactly is this? That this is the system. You see that this is the system of two point charges.

[00:00:41]Each carrying a positive charge which means that they will repel because of the force of repulsion they will move away from each other. The string will make some angle theta with the vertical and all that. Okay? Now, the question is like this if this whole system this whole system is dipped in inside a liquid. Now, it's placed inside a liquid. All right, so this whole system entire system is put inside a liquid.

[00:01:05]And it is given that the theta remains same. This angle with the string makes with the vertical it remains same. It remains unchanged.

[00:01:14]What we are supposed to calculate is the dielectric constant of the liquid. We have to calculate what will be the dielectric constant of that liquid.

[00:01:23]Okay? Wherein what is given to you is that the density of the liquid is given.

[00:01:27]And that is rho and density of spheres is also given that's equal to sigma.

[00:01:32]Okay? This question has been asked as of now four times.

[00:01:37]Three times in JEE Mains and once in JEE Advanced also.

[00:01:42]And is a very important example, okay?

[00:01:44]So, this concept may electrostatics apply hoga. Newton's laws of motion will be applied and fluid mechanics will be applied, okay? So, have a look at this problem. So, let's try to try to crack this problem.

[00:01:59]The system become place this system in vacuum. Let's place the system in vacuum. So, here I'm going to write down like this.

[00:02:08]In vacuum, in vacuum.

[00:02:11]Can you tell me one thing? Inside the vacuum, what will be this angle? This angle will be given by an equation, and what's that equation? That is tan theta equals to equals to F not where F not is the electrostatic force between two point charges when they are separated by vacuum divided by M into G. I hope that this is clear to you. I hope this is clear to you because in the last lecture we had discussed this. Okay? Also, too I can repeat repeat.

[00:02:47]One force is electrostatic, that will be in this direction, which is repulsive.

[00:02:51]Second force is vertically downwards, and what's that force called as? That's called as weight, which is M multiplied by G. This is the This is the electrostatic force of repulsion F not. I will write it down.

[00:03:04]Then, along the length of the string, there's another force what we call as tension in the string. Okay?

[00:03:10]Then, next what I can do is that what I can do is that this tension can be resolved in the rectangular components.

[00:03:16]So, you have to pass angle theta. So, it's called as T cos theta.

[00:03:21]This will be T cos theta. And this component will be T sin theta.

[00:03:28]Isn't it?

[00:03:29]Now, easily we can do one thing. We can do one thing, and that is we can do one thing, and that is we can equate this T sin theta with F not and T cos theta with mg. Reason the charged particle is in equilibrium, which means that the net force is equal to zero, which in turn means that the upward force will be equal to downward force and the force towards right will be equal to force towards left, okay?

[00:03:57]So, we have a we have a couple of equations like this. This is T sin theta equals to F naught, the electrostatic force, divided by this T cos theta is in turn equal to mg and simultaneously I have done one thing and that is I have divided the two equations. When I divide the two equations, this T and T is going to cancel. This sin theta divided by cos theta comes out to be tan theta and we have an equation that is tan theta equal to F naught divided by mg and that equation I have written down here.

[00:04:34]That is tan theta is equal to what? It's equal to F naught divided by mg. Now, I'm sure that this is clear to you and we have discussed this in the in the last lecture in the previous lecture, okay?

[00:04:46]So, let's move ahead.

[00:04:48]Now, the next what I am going to do is that we are going to dip this entire system inside the liquid. Inside the liquid, okay? So, I am writing here that in liquid in liquid.

[00:05:01]Now, is there any force which is going to change? Is there any force which is going to change?

[00:05:08]Which is going to change. So, the force number one is that the electrostatic force will change, number one.

[00:05:13]And number next is that number next is that there will be one more additional force. And that is what you call as the buoyant force whose equation is given by the Archimedes principle.

[00:05:27]Okay? So, inside this liquid inside this liquid, which force will change? Number one is that this F will not remain F naught. It will become F M which will be in turn equal to F naught divided by K where K is the dielectric constant of that liquid. Okay, number one. Number second is that in the vertically upward direction there will be one more force.

[00:05:48]That force will be bind force. That force will be bind force. Okay? So, to be to be precise, to be precise, if I'll construct the free body diagram of this object in the upward direction, there is Let me make a free body diagram of this particle.

[00:06:06]Okay? Inside the liquid.

[00:06:09]So, in the upward direction, there is one force what you call as T cos theta.

[00:06:14]Isn't it?

[00:06:16]Towards Towards right, there's a force that is T sin theta.

[00:06:21]Towards left, there's electrostatic force, which will be no longer equal to F naught. It will be rather FM. FM implies the electrostatic force inside the medium.

[00:06:31]In the vertically downward direction, there'll be mg.

[00:06:34]And in the vertically upward direction, there'll be one more force. What that force is called as? That's called as bind force. Okay? And what is the equation for this bind force? The equation for this bind force is equal to density of liquid, which is rho.

[00:06:49]Volume of sphere, which is equal to V.

[00:06:52]Into G. G means acceleration due to gravity. This is the equation for the bind force.

[00:06:58]Okay?

[00:06:59]All right. Now Now Now, what I'm going to write down is this. What I'm going to write down is this. This tan theta.

[00:07:07]Tan theta will be equal to See that See that in vacuum, what is tan theta equal to?

[00:07:13]In vacuum, tan theta is equal to F naught divided by mg. Isn't it? Inside the liquid, is this F naught going to change? Yes, of course, it will change.

[00:07:22]It will be replaced by FM. Number one.

[00:07:24]Number second is that Number second is that But this mg See in this free body diagram, mg is vertically downward direction. And bind force is vertically upwards. That means this bind force will cancel a part of this mg. Or I can say that this t cos plus fb will be equal mg, which means that this t cos theta will be equal mg minus fb. So in simple terms in simple terms, I can say that this mg will be replaced by mg minus the buoyant force.

[00:08:03]Okay?

[00:08:04]Is this clear to you?

[00:08:07]Is this clear to you?

[00:08:10]So in the vacuum, the equation becomes something like this. tan theta equals to f naught divided by mg.

[00:08:17]And in the liquid, the equation becomes like this. tan theta equals to fm divided by mg minus the buoyant force.

[00:08:27]Are these two equations clear to you?

[00:08:30]So next what I'm going to do is this.

[00:08:32]Next what I'm going to do is this. Now obviously obviously the two equations will be same. So let me write down this is equation number first. And let me write down this is equation number second. So easily I can say one thing and that is the equation number first is equal to what? It's equal to equation number second. Okay?

[00:08:50]So let me write down here that equation number first equals to equation number second. Because the reason is that their LHS same, so the RHS has to be same.

[00:09:03]Now equation number first you have that is f naught divided by what? Divided by mg. Isn't it? And the equation number second is fm divided by mg minus the buoyant force.

[00:09:18]Okay?

[00:09:19]Let me put some values here. I will replace this f naught by f naught.

[00:09:24]Then I will do like this. You know that mass you have that that's equal to density into volume because density is defined as mass per unit volume. So mass will be equal to density into volume.

[00:09:34]The mass given by me is equal to sigma into V into G because it is given in the question that sigma is the density of the two spheres.

[00:09:43]FM [snorts] will be replaced by F naught divided by K. What is this K? Dielectric constant which we are supposed to calculate.

[00:09:52]Multiplied by M M given by the liquid will be sigma V G minus buoyant force equation given by the liquid that is rho V G.

[00:10:04]Okay?

[00:10:05]All right?

[00:10:06]So so see that from this equation this F naught is equal to F naught this V G V G will cancel. So simply we are left with this that the sigma the sigma is equal to K multiplied by or I can write it like this also so that it becomes more clear to you. F naught divided by K into sigma minus rho into V into G.

[00:10:31]Okay? Now in this equation see that what cancels. This F naught and F naught is going to cancel each other. This V G and V G is also going to cancel. Which means that the sigma has to be equal to K into sigma minus rho and we are done with the problem. So that means sigma is equal to K into sigma minus rho and what about dielectric constant? So dielectric constant finally comes out to be sigma divided by sigma minus rho.

[00:11:04]Okay? So this is the final answer to this question. And I'm sure that this problem is absolutely clear to you.

[00:11:13]Okay? Now if for example in the question the value of the sigma is given the value of the rho is given you can put those two values in this equation which we have highlighted and you will get the value of the dielectric constant. Okay?

[00:11:28]That's really great. Now I can do one thing for you and that is that is let me rewrite this equation K dielectric constant is equal to D S DS means density of the sphere.

[00:11:40]divided by DS means density of the sphere minus DL DL means density of the liquid, okay? You can directly you can directly memorize this equation. I got a I got a equation directly but I have So, with a no time absolutely no time you can you can calculate what is the value of this dielectric constant. Okay? So, this problem becomes clear to you. I hope that it is crystal clear. Now, let's switch to the another example. And this example is also based on the same concept with a with a small difference.

[00:12:16]Okay? Although you got a question here, your question will be put again in NEET exam. Jee mains may be put again in exam and this question has been copy-pasted from Erodo, i.e. Erodo, okay? So, let's let's try to try to first of all have a feel of this problem.

[00:12:34]The question is we have that there are two charged particles. They are in equilibrium. All right, both are positively charged particles and they are in equilibrium. Now, see that see that the question is like this that the charges leak the charge the two particles these two particles this particle and this particle they leak out charge which means that the charge amount of charge they carry that decreases with time. Okay? At a constant rate which becomes very important, let me highlight this for you that at a constant rate this constant rate is very very important. Okay?

[00:13:12]Due to which the two spheres approach each other.

[00:13:18]Due to which the two spheres approach each other.

[00:13:23]Obviously, they will come closer to each other. Why? Because when the two charged particles leak charge, the magnitude of charge decreases, the electrostatic force of repulsion that decreases, and hence because of which they will come closer to each other with a velocity that is given to you, that is V, with velocity V.

[00:13:41]Okay. Now, this V it varies with X. So, obviously it will vary with the X, where X is the instantaneous separation between the two charged particles. As then the options are given to you. First option is X raised to the power 1/2, and second option is X, and third option is X raised power -1/2, fourth option is X raised power -1. Okay? So, let's let's let's try to solve this question, and there's an important condition given to you. Let me highlight that condition for you, that is X is very very small compared to L. A very important condition. Okay? So, have a look at these two important parameters, that's number one is that at constant rate, it leaks out charge at a constant rate.

[00:14:24]That's very important, number one.

[00:14:25]Number second is that this X is proportional X is very very small compared to L. Okay? Okay.

[00:14:33]Now, the first thing first thing I will try to do is that let me again write down that tan theta for you. And what is the equation for tan theta? Up to now, we've done three questions here up to date. So, I'm sure that now it is it is it is it is clear to you that this tan theta is equal to F divided by M into G.

[00:14:54]Isn't it clear?

[00:14:55]Okay.

[00:14:57]Tan theta equals to F divided by mg.

[00:15:00]Now, see that.

[00:15:01]See that. If this X is very very small compared to L, it is very very small compared to L.

[00:15:07]What does that mean? That means this angle theta will carry a very small value.

[00:15:14]Isn't it?

[00:15:16]This is X.

[00:15:18]So, that means this will be X by 2, this will be X by 2. when the value of this x by 2 is very, very small. It means that this theta will be very, very small.

[00:15:30]When theta is very, very small, easily we can make one approximation, and that is this tan theta can be replaced by sin theta.

[00:15:40]Equals to F, and F will be obviously equal to K into product of charges, that is Q squared divided by distance squared, which is x squared.

[00:15:50]Into m into g will remain as such.

[00:15:52]Simply I have replaced it by the I have written down the equation of Coulomb's law. F is equal to K Q1 Q2 is Q squared divided by the distance squared is x squared. That's what I have done.

[00:16:08]Okay?

[00:16:09]Now, the question arise here. Uh well, okay, sir, why have we replaced this tan theta by sin theta? See that in the diagram, the perpendicular is given to me. See that this is perpendicular, which is x by 2. It is given to me. And hypotenuse is given to me.

[00:16:25]And then you know that sin theta is perpendicular divided by hypotenuse.

[00:16:29]That is the reason. Can I write down here from this figure? Can I write down here that the sin theta, which is equal to P by H, which is equal to P by H, comes out to be x by 2L. Isn't it? x by 2L. Okay? So, let's try to put it here.

[00:16:45]This sin theta will be replaced by x by 2L. Say you have a better Okay. So, you have to make I see this is K Q squared divided by x squared into m into g.

[00:17:01]Is that clear?

[00:17:03]Is that clear to you?

[00:17:06]Now, everything is constant except this x and Q.

[00:17:11]Except this x and Q. So, can I write it like this that is x cubed is directly proportional to Q squared because I am not supposed to calculate velocity. I am supposed to calculate how velocity varies with the position, that's X.

[00:17:26]Okay? So, that's why I have written it like this. X cubed is proportional to Q squared. But you baggage I have that I have I will left that. Okay? The X cubed you have that's proportional to what? That's proportional to Q squared.

[00:17:40]Say you know, can I do like this? That is X is X is proportional to or I can write like this X raised to the power 3 by 2 is proportional to Q. Are you okay with that?

[00:17:51]Are you okay with that?

[00:17:53]Taking square root on both sides. So, this becomes X raised to power 3 by 2 is proportional to Q. Take it easy.

[00:17:59]Now, we are in search of velocity. We are in search of velocity. You should know exactly that you should know exactly that velocity and that is DX by DT.

[00:18:09]Isn't it? So, next what I'm going to do is that I will try to differentiate this equation with respect to time.

[00:18:18]Differentiate with respect to time.

[00:18:22]With respect time. Take it easy. The X raised to power N got differentiation of copper down on a chain that's NX raised to power N minus 1.

[00:18:30]To is come up with a head of X raised to power 3 by 2 here is called differentiation care I got that will be 3 by 2 X raised to the power 3 by 2 minus 1. Say yeah.

[00:18:41]Is directly proportional into into DX divided by DT which we call as internal differentiation. Take it easy.

[00:18:50]Is equal to differentiation of Q will be DQ divided by DT.

[00:18:56]Okay. Okay.

[00:18:58]Okay.

[00:19:00]I'll do one thing here. I will write down the proportionality sign instead.

[00:19:04]Is that okay?

[00:19:06]I'll see that this 3 by 2 is constant. I am not interested in that. X raised to power 3 by 2 minus 1 is X raised to the power 1/2. This dx/dt is famously called as v, which is the velocity rate of change of position with respect time is defined as velocity.

[00:19:23]is proportional to this is something constant.

[00:19:27]It is given in the question that that that the charged particles is they leak out the charge at a constant rate, which implies this dq/dt is constant.

[00:19:38]See that this 3/2 is also constant and this dq/dt is constant. I'm trying to get rid of these constants.

[00:19:44]Okay.

[00:19:45]Okay.

[00:19:47]Okay. So, I will write like this. I will write like this. This v is proportional to 1/x to the power 1/2, isn't it?

[00:19:56]Eventually Eventually I can write down this v is directly proportional x raised to the power -1/2.

[00:20:04]And finally, I'm done with the problem. This is the question. This is the answer to this problem. Okay?

[00:20:12]And I'm sure that this this question is also clear to you.

[00:20:15]Okay. It was a bit complicated, but not that that complicated. Taken G. The question you thought these two charged particles they will leak out the charge.

[00:20:24]They will leak out the charge. So, this also leaks out charge and this also leaks out charge.

[00:20:29]Because of which because of which the two particles will approach each other.

[00:20:35]Now, at at an instant, the velocity is v.

[00:20:39]Now, the question is how this v varies with the x. That was the question.

[00:20:43]And the options are given to you. So, this is the this is the correct answer that is part C.

[00:20:47]The hamne kya kiya isme? Sabse pehle hamne likha tan theta, which is f by mg.

[00:20:52]f This is what you call as Coulomb's law. This is this equation is famously known as Coulomb's law and it is an application of Coulomb's law.

[00:21:02]All right?

[00:21:03]Okay. Then next many kya kiya? I have written down this that this tan theta is replaced by sine theta. F k but they may need a liquor. Then sine theta is perpendicular divided by hypotenuse comes out to be this.

[00:21:16]Then from this equation everything is constant except this X which varies and Q which varies.

[00:21:22]So I have written simply down that X cubed is proportional to Q squared. Take a knee. Then many liquor X raised power 3 by 2 is proportional to Q. The next many differentiation here. Why I have differentiated both sides with respect time? The reason is that I am in search of what? I am in search of velocity. And what is velocity? That is rate of change of position with respect time. Which means that DX by DT is V. So I have differentiated this equation. It comes out to be this.

[00:21:53]Then V is proportional to X raised power minus 1 by 2 and we are done with this question. Okay? I'm sure that this problem is absolutely clear to you.

[00:22:03]Let's try to Let's try to work out one more one last example of the session.

[00:22:08]That's the example number four again based on the Coulomb's law. It is It is I can say that it is Coulomb's law. It is Newton's laws of motion. Okay?

[00:22:20]We can say it's a mixture of the these two. That's Newton's laws of motion and the Coulomb's law. Okay? And the question has been asked in the JEE Main 2022.

[00:22:28]Let's try to work out this example. The question is like this that there are two blocks as you can see in the figure.

[00:22:34]There are two blocks. This is one block and this is another block placed on the rough horizontal surface.

[00:22:40]Coefficient of friction mu is given to you. Mass of the block is given. Charge is given. The data is here. This is the data.

[00:22:47]Charge is given 2 into 10 to the power minus 7 Coulomb. Mass is 10 g and mu is 0.25. Okay? First and foremost I can write down here that is 10 g minus 1 g is 10 to the power minus 3 kg. So it becomes 10 to the power minus 2 kg.

[00:23:04]Isn't it? And mu is in a 0.25 to my opinion something like it is equal to 1 by 4.

[00:23:10]Okay, then 1 by 4 is also equal to 0.25.

[00:23:13]Okay?

[00:23:15]Now, what's the question? The question is like this. Calculate the value of R.

[00:23:18]R means separation between the two blocks such that the blocks are in equilibrium. Again, the blocks are in equilibrium. Again, it means that the net force on each of these two blocks is exactly equal to zero. Now, what does that mean? That means the upward force should be equal to downward force on this video block.

[00:23:40]Upward force should be equal to downward force on this block. And the force towards the right should be equal to force towards left. Once you solve that equation, you will get the value of R.

[00:23:51]Okay? And what about this R? This R is simply what? It's the separation between these two. It is the separation between these two blocks.

[00:23:59]This is the R. Okay?

[00:24:01]Now, let's focus on this one block.

[00:24:05]On this block.

[00:24:06]Now, can you tell me how many forces are there on this block?

[00:24:11]Force number one will be obviously electrostatic. Isn't it? And that will be repulsive, so I will show it like this. This will be the electrostatic force of repulsion, which is denoted by F.

[00:24:22]Okay?

[00:24:23]Uh now, since since these blocks tend to push each other away from each other.

[00:24:29]Okay? They try to repel each other.

[00:24:31]Now, because of it the force of repulsion will be there, and the force of repulsion always uh resists the relative motion between the two surfaces in contact. So, this is the direction of the force of friction. That is F. And you [snorts] should know that this F is equal to mu into n. What is n?

[00:24:47]It is normal reaction. Is there any force on this block? Yes, of course. In the vertically downward direction, there will be one more force. And what's that force is as? That is called as m into g.

[00:24:59]Okay? In the vertically upward direction, there will be one more force and that force is is is famously called as what? It is called as normal reaction.

[00:25:10]Okay? It is called as normal reaction.

[00:25:13]All right, this is the normal reaction.

[00:25:15]Okay, ji? There's one more force and that is gravitational force between these two blocks. I am going to neglect that. Because that force is negligible.

[00:25:23]That is very very small. So, let's not bother about the gravitational force between these two blocks. Okay?

[00:25:30]So, next I'm going to do Let's let's let's try to do one thing that is that is that is the upward force which is normal reaction comes out to be the downward force. What is that? That's mg.

[00:25:43]And the electrostatic force electrostatic force is equal to force of friction. So, force of friction will be electrostatic force.

[00:25:51]Force of friction is in turn equal to mu into n, isn't it?

[00:25:55]Is equal to electrostatic force equal to k q into q is q squared divided by r squared. All right? Now, this normal reaction is in turn equal to mg. Let me write down here mg.

[00:26:08]Is equal to what? It's equal to k q squared divided by r squared. Okay? And what are we supposed to calculate? What are we supposed to calculate? We will have to calculate the value of this r.

[00:26:19]So, can I write like this that this r squared comes out to be k q squared divided by what? Divided by mu into m into g.

[00:26:29]Okay? Isn't it?

[00:26:30]Now, let's do one thing and that is let's do one thing and that is substitute the values here. Substitute the values here, okay?

[00:26:39]So, r squared is equal to What is the value of this electrostatic constant?

[00:26:44]You should know the value. That is 9 into 10 to the power 9.

[00:26:48]What about q? Q is given to you. Q is given to you in the question. This is the value of the Q that's 2 into 10 to the power minus 7. 2 into 10 to the power minus 7 whole square.

[00:26:59]Because there is Q square. Take it, you see?

[00:27:01]Divided by the denominator make a mu.

[00:27:04]The mu is 1 by 4 given to you again.

[00:27:08]Multiplied by M. The mass of this block is given. What's the mass of the block?

[00:27:12]10 to the power minus 2. Let me write down here. 10 to the power minus 2.

[00:27:16]Acceleration due to gravity, let's take it 10.

[00:27:19]Okay? For mathematical convenience. The R square to I got it here will I get 9 into this 2 square comes out to be 4. 1 by 4 in the denominator becomes 4 into 10 to the power. Now bases are same, powers are added. So this is 9.

[00:27:36]Then this will be here. This is here. 7 2s are 14. And this is minus 2 plus 1 is minus 1 in the numerator. That will become plus 1. So it is 9 minus 14 plus 1. All right? Now this is here. R square is equal to 9 into 4 into 4 into 10 to the power.

[00:27:59]This is a 9 plus 1 is 9 plus 1 is 10. 10 minus 14 is minus 4. Now all these are perfect squares. So R got the value I got will I get square root of 9 is 3.

[00:28:11]Square root of 4 is 2. Square root of 4 is again 2. Square root of 10 to the power minus 4 is 10 to the power minus 2. Take it, you see?

[00:28:20]So yeah, I got R will be equal to 3 into 2 is 6 2s are 12 into 10 to the power minus 2 meters.

[00:28:29]Then we can write down the centimeters like this. It is simply 12 centimeters.

[00:28:34]So this is the answer to this question.

[00:28:38]Okay? Again, this is JEE Mains question.

[00:28:40]And a tricky question, but I'm sure that sure that it is it is crystal clear to you again. Okay? So the question was like this that there are two blocks.

[00:28:50]This is one block. This is another block. And it is given to you that both the blocks are in equilibrium, which means that the net force on each of these two blocks is equal to zero. So, first and foremost, I have identified that there are four forces on each of the blocks. One is electrostatic. Second is force of friction. Third is normal reaction. And fourth one is this weight.

[00:29:13]Okay? Then easily we can say one thing that this normal reaction is equal to weight. And this force of friction is equal to electrostatic force. Then on playing with this equation, on playing with this equation, I have calculated the value of this R, and which finally comes out to be 12 cm. Okay? So, let's stop here. See you in the next lecture.