 Hi, I'm Zor. Welcome to a new Zor education. I would like to solve one more problem related to electrostatic fields The previous lecture was in my personal view quite remarkable in the fact that the result was I would say counterintuitive We had an infinite plane charged with some Electric charge and we were measuring the intensity of the field and We found basically that at any point in space regardless of the distance to the to the plane The intensity will be the same Today's problem is well in some way similar and Again, the result is something which I would consider a little bit counterintuitive So here is the problem We will consider an Empty Sphere very thin. This is the center. So this is the sphere So it's a thin surface empty inside and it's charged with some Density sigma which means that any area of this Surface of the sphere Has a charge proportional to the area itself and the coefficient of Proportionality sigma to get coulons. Let's say from Square meters now there is a point inside that sphere at distance H From the center So this is a point P and the distance H is to the center Okay, and I would like to measure the Intensity of the field in this particular in this particular at a particular location Now I will do exactly the same as I did in the previous Problem I Will use certain pieces of this sphere which are The same in terms of their distance to the point P is the same and The angle of the vector from every point of the surface which I'm considering To the point P has the same vertical component. Now. What is it? What is this piece of this sphere which I'm talking about well consider? So this is the vertical z-axis, okay now consider you have planes parallel to x y plane At the distance R from a center, so it will be something like this Now I will have two planes very close to each other One is at a distance R from a center and Another is at a distance R plus dr Where dr is infinitesimal increment of this distance. So this piece is R Now what is Between these two surfaces Between these two planes on a spherical surface. Well, it looks like a ring basically, right? It's not exactly the ring because the radius is are slightly different way are slightly different But anyway, we can call it a spherical ring now what's Good about this ring is that from any point on this ring The distance to point P is exactly the same Let's call this point a for instance right on the right side edge of this spherical ring so What we can say that the radius is exactly Not not radius the distance is exactly the same and the projection of this distance on to The z-axis and the vertical is this piece is always the same to right Now what does it mean? If I will take an infinitesimal piece here and here and here and here and here anywhere on this ring If these infinitesimal pieces are of the same area, which means of the same charge They will have The vertical component of the intensity vector Exactly the same why because from this let's say intensity vector depends on this charge and The distance right remember the Coulomb's law We have intensity is equal to k times q divided by L square well L the distance and q is this charge Well, actually, it's a small one differential actually but in any case This is the intensity so L is the same because that's the same distance Q is the same these are all infinitesimally small pieces of the of the ring and When I'm talking about the vertical component. It's basically the vector times Sign of this angle right so if this is the vector let's say This is the force for instance. This is positive and this is positive So it's a repelling force now. What's the vertical component of this? I have to multiply the force by the sign of this angle and The angle exactly the same for all points on this ring so Vertical components can be added together because they're all Pointing along the z-axis. How about horizontal components of the same vector this one well, all horizontal components will nullify each other because for every piece here I will find the opposite piece on the ring which has the opposite direction of the horizontal Of the horizontal force Now the the vertical will be exactly the same as this one. So all the verticals are Can be combined together by integration and all the horizontal can be ignored because they will nullify each other For every point there is an opposite one Okay, so what we have to do is we have to calculate the total amount of Electricity in the ring from this we can calculate the magnitude of the intensity Force of intent intensity force and take the vertical component and then integrate Okay, okay, so that seems to be relatively simple plan first of all, what's the charge of the entire ring? That's the surface of the ring area of the surface of the ring times Density okay, now what is the surface of this ring now from Geometry, we know that And you can by the way go to mass 14 scores on the same website. It's all explained there. So 2 pi r h is the formula for formula for a dome so if you have A sphere and you have a dome top part of it and If the h is the height of this dome then this is the formula for a surface of This dome now to get the ring we have to basically take one dome Which is based on a lower one then another one based on the upper Boundary and then subtract them right so the bottom one Has a height r minus a lower case r So h is equal to r minus r Let's go to h1 h2 is equal to r minus minus r Minus dr, right So if this is distance r for the lower and then r plus dr for the hot for the for the upper cutting plate then the difference between them is this one So my area is equal to 2 pi r times h1 minus h2 Which is actually dr. So my surface area is equal to 2 pi r dr Now if this is my surface area of the ring then the charge in the ring is equal to times sigma and this is a differential of Q where Q is the charge of the entire Sphere so it's just the charge in this ring Now if I have the charge I can use this formula to calculate the intensity Vector magnitude of the intensity vector and it's equal to I Will also use D here because it's only for this ring. So it's differential and it's equal to this Chi K times this 2 pi r Sigma dr Divided by L square. Okay. What is L square L square is the distance between P and and a Well, let's just think about I can use this triangle I need this Catatouille's and this catatouille's now this catatouille's I can use this triangle So if this is the point B, let's say then the distance between O and The plane is R The distance from O to A is radius so I can find AB AB square is equal to R square minus R square So I've got the AB square now BP Square Now just BP BP is equal to H minus R So a P square is equal to A P square is equal to BP square plus a B square which is H minus R square plus R square minus R square Let's simplify it a little bit. I will have R square. I will have H square. I Will have minus HR and R square with minus R square will cancel each other. So this is my L square And this goes to a denominator Okay, got that. Now I have to find the vertical component of this of this vector So the magnitude is this one Now all these vectors are Differently directed, but their vertical component is the same and that's why we have to really That's why we can integrate the whole thing Okay, so what's the vertical component? I have to multiply this vector by sine of this angle sine of this angle is H minus R divided by L So if I will use the Z Not Z not not here Z here Z So which means a Z component vertical component I have to multiply it by the sine which is H minus R divided by the same Thing but the real length not length square but real length which is square of this Which means the whole thing will be in the power of three seconds, right? and This is something which I have to integrate by R from minus R To plus R Now the details of the of the integration. I'm not going to to have here I would refer you to the notes for this lecture at unizor.com It's really a simple integral all you have to do is substitute the whole thing in in parentheses here as x and This is just a linear function of R. So it's just linear transformation which will give you much simpler Expression for X very easily integrated. Basically. It's a power function no more than that and I will write down the answer to this So the answer of integration and again, you can take the details from the website is the following Now I have a constant C. I just all these constant I will put together and Will not bother about it. You see all these little things and now in these terms the integral is equal to to C times R square minus h squared divided by square root of x plus Square root of x Where x is this and I have to substitute this Since I changed the x r to x. I have to change the limits of integration so limits of integration will be from R minus h square to r plus h square Okay now So this is the answer to integration of this by x where x is this expression and That's why I put instead of r. I put expression in terms of x and instead of dr correspondingly expression relative to dx now What is this? Well, we have to substitute instead of x this and Then minus substitute this for x. That's formula of Newton-Ladenitz formula, right if you have a Indefinite integral then definite integral would be an indefinite if you will substitute corresponding values. Well, let's do that so again x is R square plus h square minus 2 h r That's how we substitute it, right? All right, so let's instead of x substitute r plus h square of x would be r plus h So it's 2c times R square minus h square is r minus h times r plus and it will Cancel this one. So I will have r minus h plus r plus h That's the upper limit the lower limit will be and here we have to be very careful Our point is inside which means h is less than r H is less than r that what that's what it means inside because h is the difference It's a distance actually between the point and the center and r is the radius So if that is true, then Square root of r minus h square would be r minus h because this is a arithmetic value of the square root So I have to subtract r minus h square divided by r minus h, which is r plus h plus Square root of r minus h r minus h And what will I have? H H H H Now r and r and minus r and minus r zero and This is what I was talking about about remarkable results Intuitively you might actually think that if you are inside a sphere if you're closer to the surface from inside You might actually get more Force from from this side where you're closer to then from the opposite No, it's exactly the same and nullify each other the whole integral is equal to zero So the intensity of the force is equal to zero there is no So if you place any charge inside a charged Sphere it will not experience any kind of a force on it There is no repelling and no attraction to the sides of the sphere It will be completely neutral as if sphere doesn't have a electricity at all What happens if our point is outside of the r outside of the sphere Okay, let's do exactly the same thing, but now square root of r minus h square Would be equal to h minus r right because we have to have the positive value So what will be is to see Now r minus h Square square minus square now with a plus will be fine plus would be r plus h So it will be r square minus h squared divided by r plus h. So it's r minus h Plus R plus h so that's my the same thing as in the previous case But with a lower limit would be slightly different so minus now The square root of x in this case is h minus r So if this is r square minus h square we divide by h minus r. It will be minus r minus h right r square minus h square divided by r minus By h minus r is Minus r square minus h square divided by r minus h is Minus r plus h, right? So that's why I put minus r minus h Now plus square root is h plus r Minus r sorry h mine h minus r So H goes out H goes out But now r you see r and r and this is minus and minus and minus so it will be four hours So it will be eight c times four r Which is equal to By eight to see sorry it will be If I multiply Now what is my C? C is my so it's 2 pi K Sigma r Divided by 2h square, but now we have 4r so it will be 8 Square Or 4 Pi r square 4 pi r square Sigma K Divided by h square and this is very interesting formula. Let's just think about what it is 4 pi r square is the total surface of the sphere as we know from geometry Time sigma is the total charge because it's uniformly charged K is a Coulomb's constant now h is the distance to a center of a sphere So this formula is exactly the same as if the whole Charge concentrated in the sphere would be positioned in its center Only because this is now the distance to a center so it will be the total charge divided by The distance to a center so as if the total charge is concentrated in the center so a sphere acts on any outside object as If the total charge of a sphere is concentrated in its center You probably understand that it's kind of relatively easy to prove that if it's not an empty sphere But a solid sphere which obviously can be represented as as many Concentrical spheres infinitely thin each one and for each one the total charge The total tendency intensity would be as if it's a total charge concentrated in the center Then we can summarize them all together because all of them are acting in exactly the same way the total charge divided by Square distance from a center. So the first solid Sphere for a ball filled with electricity uniformly filled with electricity. We will have exactly the same story And that's the conclusion. So the empty sphere inside has no electricity Outside it's the same as if the whole sphere is basically reduced down to a point Where the center is located with the total charge That's it. Thanks very much. Please do pay attention to the notes on this website for this lecture Because it contains all these integral derivatives, etc. So thanks a lot and good luck