4. (a) An inverted hemispherical bowl of radius R carries a...

4. (a) An inverted hemispherical bowl of radius $R$ carries a uniform surface charge density $sigma$. Find the potential at point $a$, the centre of the hemisphere and at point $b$, the "North pole". [9 marks] (b) A point charge $q$ is placed at the origin and is surrounded by a cylindrical surface of height $2h$ and radius $R$ such that the cylinder axis coincides with the $z$-axis and the surface extends an equal height above and below the $xy$-plane. This is shown in the figure below. (i) Find the flux passing through the sides of the cylinder only. You may find the following result helpful: $$int frac{dz}{(R^2 + z^2)^{3/2}} = frac{z}{R^2sqrt{R^2 + z^2}}$$ Hint: Convert the electric field into cylindrical polar co-ordinates first. (ii) As $h o infty$, how much flux passes through the sides of the cylinder? Is this consistent with Gauss's law? [16 marks]

Transcript

00:02 In this question, we are given uniformly charged sphere, on radius of the r is being placed inside an uncharged cylinder.

00:18 Okay, so there are center coincides.

00:23 So the radius of the cylinder is big r, the length of the cylinder, is l, then it's greater than 2r, and then there are five parts in the question.

00:40 So we want to find the electric flux through the curve surface, the top circular cap, the bottom circular cap, find the total flux, and check that with, and compare the result with gauts's law.

00:56 Okay, so to solve this question, okay? for the curve surface.

01:03 Okay, this is the point we will look at.

01:08 Choose the point down here.

01:15 So the electric field will be pointing like this.

01:19 Okay, and then the area that we are considering is going to be here, okay, on the curve surface, which is given to be 2 pi r, the r, to buy r d z sorry and i'm going to also set up a coordinate system you should look something like this okay so this is our x and then this is our z the y will be like that okay so this is our z and then this is our p -r, okay, so this distance here, like i'm just going to call it like r prime okay, r -prime by pythagoras theorem, okay, from diagram r -prime square is equal to z square plus r -square and then the e at r -prime, is going to be q divided by 4 pi epsilon not r prime square.

02:47 That's a magnitude.

02:54 Okay.

02:56 And then when we consider the flux, the electric flux is equal to integral of e.

03:03 That's the surface integral of e .d .a.

03:07 And this is e cosine data da.

03:11 So we actually need that expression of so the cosine data, okay, so the cosine data on the diagram will be this angle data, okay? so it will be this angle data.

03:25 So, cosine data, okay, from the diagram, it's going to be r divided by r prime, is r divided by square of z squared plus r big r square, okay? and we are given that da is 2 pi r dz so we can actually do our integration so phi e is integral of e cosine data ecoside data d .a so put in all the expressions t over for pi x or not and then with the cosine data i get r divided by z squared plus r square plus r square q over 2 and then 2 pi r d z so you can take out the constants you get q r square over 2 f0 and you are left with the integration is along d z so minus l over 2 to l over 2 okay so the integration okay what we are left with in the integral is 1 over r square plus z square 3 over 2 d z and then you check the table for the integration results at the back of the book okay this is what you get okay one over r square 1 over square root z square plus r square and then you have the limits l2 and negative l2 and there's a z on top okay so this is why you add up q over 2 2 x0 the r square cancels okay l over square root l over 2 square plus r square okay, so this is the answer for part a.

05:47 Okay, next for part b, okay, we want to find the electric flux on the top cap.

06:06 Okay, so i'm putting my drawing.

06:15 Okay, we have a sphere inside here.

06:18 Okay, so the da will be on this surface, from this point, the e will be pointing like this, and then the da is pointing up.

06:40 So we need the cosine data.

06:44 The data will be here.

06:48 And so, and this is our l over 2, and this is our small r.

07:00 So again, gauss's law, no, this is not gauss's law, just finding electric flux.

07:11 A is equal to integral of e .da, d 'a, integral of e .cci data, da...

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