11th Class Physics Chapter 9 | Numerical Problems 9.1 and 9.2 | 11th Class Physics New Book 2025

[00:00:06]Assalam Walekum Dear Student This is I wish Madi your Physics Teacher Hopefully you all are well student today we have to discuss numericals of chapter nine and let's move towards the statement of numericals which you have discussed some formulas in this chapter so far and I had also highlighted to you that we have to use these in numerical problems.

[00:00:27]First you had the formula for Coulomb's Law.

[00:00:29]f = kq1 q2 over r² Then for the electric field you used.

[00:00:36]You used the electric field as a potential gradient. Ok? And that's how you dealt with electric force.

[00:00:41]Similarly, we will also highlight whatever formulas there are.

[00:00:45]Ok? So the first statement you have is that two identical point charges repel each other. There are two opposite point charges. What are they doing? Repelling each other.

[00:00:55]Ok? Suppose this is your first point charge Q1, this is your second point charge Q2, and what are they doing to each other? They are repelling.

[00:01:05]Ok? Meaning the charges are same. Both are positive charges. You will say that the force that they are applying between them, what is the value of that force? 0.4 Newton. And the distance between them is 5 cm. Is.

[00:01:21]Now what is the value of this distance that you said? 5 cm Now when the distance is 5 cm then the force applied is 0.4 Newton. Second, he says that if we make this distance 0.5 cm. to 2.5 cm. If we do it then how much force will we apply on each other? And in the second scenario it says that if you make this distance again 15 cm. Ok? How much is the first? 5 cm So how much force is being applied? 0.4 Newton. And you reduce this distance to 2.5 cm. Then how much force will be applied? We have to find this out. And if you change this distance by how much? 15 cm Then how much force will be applied? So in these two steps you have to do both these calculations that first when the distance is 2.5 cm. Then how much force will it be? And second is when you have a distance of 15 cm. So what do you have to do then? The force has to find out. So this is your statement. Now student, you know that here you will use the inverse square law to find out what is force?

[00:02:27]With the square of the distance in an inverse relationship with the distance.

[00:02:30]Ok? And look at the rest of the things, he is just varying the distance. If the distance is very, then the force will be very. Ok?

[00:02:36]So what does this mean? So if you talk about ah Coulomb's law f = kq1 q2 over r² then these three things will be constants here. You already have the constant value of k. The charges q1 and q2 are fixed. Ok? The magnitude of the charge is not changing. This means in both cases we will keep q1 q2 constant.

[00:02:57]If both of these are constants then what does it mean? What is K Force? Inversely proportional to the square of the distance.

[00:03:05]Ok? So you can simply write this as f is inversely proportional to 1 r².

[00:03:12]Next, you can create this same form.

[00:03:15]Ok? Or you can write it in this form.

[00:03:17]You can do it both ways.

[00:03:18]And again in this form, if you have Force F1, what is the distance? R1. Ok? So you will put these values ​​here R1 and F1. Now if you simplify it and write the constant, it can be written like this.

[00:03:32]Now there you have equation number one.

[00:03:34]What do you have? Equation number two. You have to divide these two.

[00:03:38]When divide f over f1 okay? You divided the left side from the left side.

[00:03:44]And when you divide the right side from the right side, you will have KQ1 Q2 and KQ1 Q2 will get cancelled.

[00:03:52]Ok? And simply the relationship will be formed. If we look at this, let me give it to you in the rough. Here you will see that f over f1 = kq1 q2 over r² by k q1 q2 over r1². So students, you can simply flip it and write it on top. When you convert this division into multiplication, this value will go up to you.

[00:04:24]r1² over kq1q2 If you cut this from this, what will simply come out? F over f1 = r1² over r² This will give you the mathematical form. Ok? Now you have to find out f1. Cross multiply it. f² = f1 r1² From here you can find f1. This value will be divided on this side of yours. So f1 = fr² over r1². So this is our final mathematical form, which you can use to find the force. Now we will put value in it.

[00:05:00]How much force do you have? 0.4 Newton and distance is 5 cm. So you will convert centimeters to meters because what do you have for SI units of distance? There is a meter. So what do you do about it? Let's divide it by 100. We will divide by 100. This value will come to 0.05.

[00:05:17]Put this and how much R1 do you have? If you divide 2.5 2.5 by 100 then the value will come to 0.025.

[00:05:27]Take its square also. Take its square also.

[00:05:29]Multiply by 0.4 and divide by the square of this value. So how much will the f1 you have come to? 1.6 Newton will come.

[00:05:36]Similarly, now students, you can change this mathematical form that you had created like this, f1 is equal to f r² over r1², so if f1 is there then it means here r1 is r, if I make it r f2 here then this distance will be how much, it will become r2, so similarly you can write f2 = f r² over r2², okay, so you can simply use this directly, if you want you can also calculate it again.

[00:06:09]In this case, your equation number one will be the same.

[00:06:11]But what will come into this equation here? F2 and what will come here? R2 will arrive in a matter of seconds. So you can also direct this much.

[00:06:22]You can use this relation directly. Then put the value of F2 equal to F.

[00:06:28]Put the value of R. Put the value of R2.

[00:06:30]And so how much R2 do you have? 15 cm What will you divide by to convert this into this meter? 100, then you will get this value 0.15, take its square also and divide it by the square, then what will be f2 equal to? The value will be 0.04 Newton. So in this way you have found both the forces and what is the value of force you have after changing the distance? There has been a change.

[00:06:57]And you found both. Now friends let's move to numerical number 9.2, first let's understand its statement. A particle of charge 20 microcoulomb. There is a charge. What is the charge on a particle? G20 Micro Coolam. Now what is the value of micro? 10 power -6. So you will get the total value of the charge, what is that? 20 * 10 power -6 will come to Coolam.

[00:07:25]See next. In this you have that is placed between two parallel plates 10 cm. Apart and having a potential difference of 0.5 k. Now what is the potential difference between these two? 0.5 KV. Now what will you do with the kilo? Like here the prefix used was micro. And what is the value of micro that I put? 10 power -6. So what is the value of a kilo you have? 10 to the power 3 10 to the power 3 is basically 1000. If we multiply this by 1000 then it will become 500 volts. What is the potential difference and the distance between these plates? 10 cm Now if you convert this into meter 10/100 then this value will come to 0.1 meter. Now what you have to do is find the value of the electric field.

[00:08:13]Basically you have to find out the electric field intensity and secondly you have to find out the electric force.

[00:08:18]Ok? So you have these two things that you have to find out.

[00:08:22]What will you do first? To find out the electric field intensity, we will use the formula for electric field as a potential gradient. E = V over D. Right? Now V here is the potential difference. D is the distance between the plates. The value of V is also given. The value of D is also given. How much value of V do you have? How much is 500 and D? 0.1. We will divide 500 by 0.1. This will come to 5000.

[00:08:46]And what units do you have for electric field? Meter on volts. You can write like this also. And do you also know what are the second units of electric field? Coolum on Newton. So you can write any of these two units.

[00:08:58]Ok? And if you see the answer in the book, he has written it in the form of 5 Kilton per Coolum.

[00:09:04]So you can write this 5000 like this also.

[00:09:08]5 * 10 to the power of 3 and coulomb to the newton.

[00:09:14]Ok? So this 10 power 3, you write it as kilo prefix and what will you get next? Coolum will arrive at Newton.

[00:09:21]Ok? So this way you can find its value of electric field.

[00:09:25]Now the second is that you have to find out the electric force also.

[00:09:27]So from the formula of electric field itself you had derived the electric force that e = fe over q force on unit test charge. So what will fe equal to be? q * e then what is the value of q?

[00:09:41]Charge 20 * 10 to the power -6 and what is the value of e you have now? Multiply 5000 and this will give 100 * 10 power -3 Newton.

[00:09:52]Now if you write it like this then this is your correct answer. But if you look in the book, it is written in Milli Newton's form. So what should you simply do? You can also write 10 to the power of 3 in milli form.

[00:10:02]So see what different prefixes you can use. Even then you have to do the calculations. So one thing to keep in mind is that the value should be in SI units. If there is distance then you will take it in meters. If you have the value of the charge then you have to take it to Coolam.

[00:10:16]If you have the value of the electric field, you can also see its units.

[00:10:20]You have to take the force in Newtons.

[00:10:22]So if you do not have given in this value then you have to put these prefixes and move to such units. Ok?

[00:10:37][Music] [Music]