2. Let's return to the charged ring from Homework...

2. Let's return to the charged ring from Homework 2: a dl P Recall that this ring had a charge density of \(\lambda\) C/m, radius a, and a total charge Q. (a) Determine the electric potential V at point P. (Hint: Use a similar approach as to when we found the electric field of this ring!) (b) Using your answer from part (a), determine V at the origin. How is this different from the magnitude of the electric field at the origin? (You may use the expression for E from Homework 2, and need not re-derive it.)

Transcript

00:01 In this question, we have an annulus, which looks like this, inner radius a, outer radius b, and then this is our coordinate system, x and y.

00:21 Okay, so a is 5cm, b is 10 cm.

00:30 We want to find the electric potential along the xxies it has a charge density sigma after then we want to find a e -few and then take the limits so that it looks like so that we get the result for infinite thin sheet of charge and calculate the potential at origin given the some values and find the final speed if you release a charge from the center of the annulus.

01:11 So to do this question in part a, you want to find the electric potential.

01:21 Okay, so we'll be using dv equals to eq over 4 pi epsilon not r.

01:30 Okay, so this is the point of interest.

01:35 And then we consider block of dq over here so the distance r is going to be r square plus z squared square root okay so this is how i'm going to write this out and then the dq okay so make it different, i'll just put x here.

02:11 Okay, so from the diagram, x is square root of r squared plus z square, and then the dq is equal to sigma d a, and then this is sigma times two pi r, d r, okay? so v is integral of dq over 4 pi epsilon r just put in the terms sigma 2 pi r d r d r divide by x sorry okay x here and then we have 4 pi x x not square roots of r plus z squared and the integration is from a to b.

03:11 So bring out the constant sigma over 2, f s .0, integrate, okay, from a to b, r the r, d r over square root of r squared plus z square.

03:30 Okay, so this one can be integrated easily.

03:36 So it's just square root of r square plus z square.

03:39 A to b and so we get r square plus e square root minus r square plus a square square root okay, so this is the answer for part a okay then in part b we want to find the electric field at all points along the x -axis by differentiating the potential so e equals to negative dv d z okay and you get sigma over 2 aft not z over square plus b square minus z over square of i wrote something wrong over here okay it should be b square plus z square square plus z square okay yeah that should be the answer so you have a square plus z square okay that's how it should look like okay after you differentiate the potential respect to z okay but make sure you know you put the negative sign okay so talking about that i need to put a negative sign here okay, so in part c, taking the limit, a to 0 and b to infinity, okay.

05:39 So this is what we get for a to 0, which means that the term with a, a goes to 1, and then b towards infinity, means that the term will be go to 0.

06:00 So the e will be negative sigma over 2 x0x00...

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