Summary Lecture 01

We are going to review the entire lecture series. We started with the rule of Kulam. The rule of Kulam is that if there is a distance r between two waves, then the force applied on them is 1 by 4 pi epsilon 0 q1 q2 by r square. This is a satish rashi. If the charges are of equal magnitude, then the force is repulsive. If the charges are of opposite magnitude, then the force is of attraction. The most important idea we had was the idea of electric field. According to this, if there is a force q, which we consider to be point force, then 1 upon 4 pi epsilon 0 q by r square. This is an example of an electric field in a magnet. The thing we started to learn is that can we calculate this electric field easily if we have a distribution of impulses, a charge distribution. For that, we had given a formula that if there is a charge distribution rho r prime, this is a coordinate system and rho r prime is the density of the impulse, then 1 upon 4 pi epsilon 0 rho r prime divided by r minus r prime cubed r minus r prime. This is a static electric field and it does not depend on time. For this, we had shown that its divergence is rho r upon epsilon 0 and its curvature is zero. We can show that we can write E as a gradient of a particle. E is a conservative field. If an impulse is taken from one point to another, then the work done does not depend on its direction. Let us spend some more time on this. So, according to Gauss's theorem, divergence E is equal to rho r upon epsilon 0 and if we use the divergence equation, we can show that E dot ds is equal to Q enclosed by epsilon 0. This means that if I apply E dot ds on a closed electric field, then I can write E dot ds as delta S. If I apply E dot ds on a closed electric field, then I can write E dot ds as delta S. So, this integral E dot ds is equal to Q enclosed by epsilon 0. We call this integral form of Gauss's law. Now, we have written del divergence of E is equal to rho r upon epsilon 0. We call this differential form of Gauss's law. This is an electrostatic field. Since this is the last week, I will mix some English in this. Let us see this in both differential form and integral form. Let us see this in both of them. This means that there is no excitation left on the electrostatic field. This means that E is equal to 0. We have found that there is no excitation left on the cavity of an electrostatic field. We can write E as a potential or gradient. So, the gradient is minus rho r upon epsilon 0. We call this Poisson's semicircle. We have also found that the calculation of Vr is easier than the calculation of E. So, the complete electrostatics is based on the calculation of Vr. We call this as potential theory. If there is no excitation at any point, then del square V is equal to 0. We call this as Laplace. For example, if we have a point charge, its electric field looks like this. If we take this region, then V is not equal to 0, E is not equal to 0, but del square V is equal to 0. And there is no maximum or minimum in the solution of del square V is equal to 0. And the result of this is Unchoss theorem. We have also seen this. So, we did some experiments with this. After learning about E and Vr, we calculated the energy contained in an excitation-dispersion charge distribution. We will write its mass as 1 by 2 integral rho r rho r prime divided by r minus r prime dV dV prime. We can also write it as 1 by 2 rho r Vr dV. But the main thing we have learnt from this is that, that we can consider the energy contained in an excitation-dispersion charge distribution as 1 by 2 integral rho r rho r prime divided by r minus r prime dV dV. Again, as I said earlier, that electric field or electric field is not just a convenient way to see a force, but it is a sign which we can see, which we can feel, which we can feel. So, to think that an energy is contained in the electric field, is a very important concept. So, if I want to find out the energy, I can do it in this way, that I take epsilon 0 divided by 2 E square at every point and integrate it over the entire volume. So, this also gives us an energy. Through this we have found the meaning of energy contained in various electric fields and in excitation-dispersion charges. The expectation that you will have in this relationship is that you can learn about the electric field and its forces. a the 3. You have learnt about the solutions of del square V equal to 0 and del square V equal to rho R upon epsilon 0. 4. You have learnt about the uniqueness of the solutions of del square V equal to 0. 8. You have learnt about the uniqueness of the method of images.