A sheet of infinite moving charge and a point charge A very...

A sheet of infinite moving charge and a point charge A very large metal sheet has charge moving over it. The sheet has a surface charge density of $\sigma$ = 5x10$^{-6}$ C/m$^2$. The charges on the sheet are moving at an average speed of $v_s$ = 0.0001 m/s. A proton is located a distance D = 10 cm from the sheet. The proton is headed straight at the sheet with a speed of $v_e$ = 10000 m/s. a) What is electric field above the sheet at the point the proton is at (magnitude and direction)? b) What is the surface current density of the sheet? c) What is the magnetic field above the sheet at the point the proton is at (magnitude and direction)? d) What is the electro-static force on the proton (magnitude and direction)? e) What is the magnetic force on the proton (magnitude and direction)? f) How does the electro-static force compare to the magnetic force (which is bigger and by how many times bigger is it)? g) Approximately how close will the proton get to the plate? (is there a force you can neglect?)

Transcript

00:01 According to this exercise we've been given to very large charged sheets with same and opposite charge density.

00:09 The one on the left has a charge density of sigma -not and the one on the right has a charge density of negative sigma -knot.

00:21 Now the key point here is that it says very large charge sheets meaning a separation between the sheets d, it's much, much more about any of the other dimensions of the sheets like their width or their height.

00:39 Yeah, that will be useful later on.

00:41 We have multiple tasks for this exercise.

00:44 The very first one asks to draw arrows at each of the four given points, nbcd, for the direction of the electric field at these points.

00:58 Immediately to say that the direction of the electric field and the magnitude of the electric field at all these points at all these four points is exactly the same why because this is the exact configuration of a capacitor same and opposite charge density on the two yeah on the two sheets therefore the electric field inside the between the air -to sheets exactly the same along all over the space between the sheets therefore yeah we can argue like this that yeah the electric field is exactly the same at all at all points and okay we will we will prove the formula for the magnitude of the electric field due to a single very large sheet which is sigma not over two epsilon not we will prove it at in task d, for the moment, okay, let's take it for granted.

02:04 Let's use the superposition principle.

02:07 Let's use the superposition principle, meaning that at every point, any of the four points, we have two contributions, two separate contributions due to the two charts sheets.

02:24 Okay, and both they point to the right, to the positive.

02:29 X direction.

02:30 Here i draw a right -hatted coordinate system.

02:34 Okay, they point towards the right.

02:36 That's why i'm using the x -hat notation here.

02:40 And okay, the magnet is for both sigma knot over two epsilon not.

02:48 Therefore, the total electric field at each of the points is simply a sigma -0 over epsilon knot x -hat.

02:59 Here we can argue like this we believe we don't have to expand any further let's go to the next task so we we assume a path from a from point a to point c and we've been asked about the work now what is the definition of the infinitesimal work is the dot product between the force accepted dotted with the infinitesimal displacement d l okay the electric field electric force equals a charge times the electric field.

03:35 Okay.

03:37 The displacement here, the infinitesimal displacement.

03:41 For the specific part, always points down.

03:46 Okay.

03:47 Let's say that this deal.

03:49 Always points down.

03:51 And it is perpendicular to the electric field anywhere along the path.

03:57 Anywhere around the path.

03:58 Therefore, the door product.

04:00 Is just zero.

04:02 And therefore, the infinitesimal work is zero.

04:05 At any point of the path.

04:10 Therefore, the finite work, the total work done is zero.

04:15 This is the answer.

04:16 This is the answer for b .i.

04:19 Now, similarly, for b2, this is the formula for the potential difference from a to c.

04:30 Negative, we always need to remember that there's a negative sign here, negative the line integral of the electric field along the path.

04:38 Okay, this is quite demanding mathematically anyway.

04:45 But if we remember that the electric field is always perpendicular in this specific case with the infinitesimal displacement along the path, therefore again, this potential difference is zero as it was for the work.

05:01 Okay let's go to the next exercise ci this time we have a path from point a to point d from point a to point d mm -hmm and this time the infinitesimal displacement is not it's not always perpendicular to the electric field and we've been asked about there again for the work and the potential difference okay we're using again the the formula for the infinite work force exerted dotted dotted with infinitesimal displacement we're doing this trick that we usually do in this kind of exercises we we split infinitesimal displacement into two components a component that is parallel parallel to the electric field and a component that is perpendicular to the electric field always in a vector sense anyway and if we okay and if we express infinitesimal displacement like this and if we do it again with the electric field therefore it's easy to see that the component that is perpendicular to the field goes away just gives zero the dot -product gives zero and we are left with an expression like this electric field dotted with the component of the infinitesimal displacement that is parallel to the electric field and since this dl parallel has exactly the same direction, that's exactly the same direction, it points to an exactly point of direction to the right, as with the electric field, therefore, the dot is a positive quantity, therefore the infinitesim work is a positive quantity, because q is also positive, and eventually we say that the work is positive.

07:05 Similarly, for the potential difference, we're using again the definition and the formula for the potential difference...

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