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4. (100 points) A very thin plastic rod of length L is rubbed with cloth and becomes uniformly charged with a total positive charge Q. In this problem you will be asked to calculate the electric field a distance d from the top of the rod and on axis with the rod at observation location "*" as indicated in the diagram. A. Determine an expression for the amount of charge dQ on the infinitesimal piece dy in terms of the known variables. B. Derive an expression for the vector electric field d?–‹E‹ of the infinitesimal piece dQ of the rod at the observation location in terms of the given variables and known constants. Sketch, on the diagram, the direction of this field. C. Integrate over the charge distribution to determine the electric field vector at the observation location. Your answer should only contain the given variables L, Q, d and known constants. D. What is the field in the limit the observation location is far from the tip of the rod. Your answer should only contain the given variables Q, d and known constants.

          4. (100 points) A very thin plastic rod of length L is rubbed with cloth and becomes uniformly charged with a total positive charge Q. In this problem you will be asked to calculate the electric field a distance d from the top of the rod and on axis with the rod at observation location "*" as indicated in the diagram.

A. Determine an expression for the amount of charge dQ on the infinitesimal piece dy in terms of the known variables.

B. Derive an expression for the vector electric field d?–‹E‹ of the infinitesimal piece dQ of the rod at the observation location in terms of the given variables and known constants. Sketch, on the diagram, the direction of this field.

C. Integrate over the charge distribution to determine the electric field vector at the observation location. Your answer should only contain the given variables L, Q, d and known constants.

D. What is the field in the limit the observation location is far from the tip of the rod. Your answer should only contain the given variables Q, d and known constants.

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4. (100 points) A very thin plastic rod of length L is rubbed with cloth and becomes uniformly charged with a total positive charge Q. In this problem you will be asked to calculate the electric field a distance d from the top of the rod and on axis with the rod at observation location "*" as indicated in the diagram.

A. Determine an expression for the amount of charge dQ on the infinitesimal piece dy in terms of the known variables.

B. Derive an expression for the vector electric field d?–‹E‹ of the infinitesimal piece dQ of the rod at the observation location in terms of the given variables and known constants. Sketch, on the diagram, the direction of this field.

C. Integrate over the charge distribution to determine the electric field vector at the observation location. Your answer should only contain the given variables L, Q, d and known constants.

D. What is the field in the limit the observation location is far from the tip of the rod. Your answer should only contain the given variables Q, d and known constants.

Added by Thomas M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Jacob Fry

06/13/2022

Step 1

Given that the rod is uniformly charged with total positive charge \(Q\) and has a length \(L\), the charge density \(\rho\) is \(Q/L\). Therefore, the amount of charge \(dQ\) on the infinitesimal piece \(dy\) is: \[dQ = \frac{Q}{L} dy\] ** Show more…

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4. (100 points) A very thin plastic rod of length L is rubbed with cloth and becomes uniformly charged with a total positive charge Q. In this problem you will be asked to calculate the electric field a distance d from the top of the rod and on axis with the rod at observation location "*" as indicated in the diagram. A. Determine an expression for the amount of charge dQ on the infinitesimal piece dy in terms of the known variables. B. Derive an expression for the vector electric field d –‹E‹ of the infinitesimal piece dQ of the rod at the observation location in terms of the given variables and known constants. Sketch, on the diagram, the direction of this field. C. Integrate over the charge distribution to determine the electric field vector at the observation location. Your answer should only contain the given variables L, Q, d and known constants. D. What is the field in the limit the observation location is far from the tip of the rod. Your answer should only contain the given variables Q, d and known constants.

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Transcript

00:01 So in this question we're being asked to calculate the electric field due to a rod which is uniformly charged.

00:09 So this is our situation here.

00:11 We call a rod.

00:13 And the length of the rod is l and it has a total charge q on it, which means that the charge density, row q, is q divided by l because it's uniformly charged.

00:27 Now we're looking at an observation point, a distance d above the rod.

00:37 And so first of all, we want to think about the charge dq on an infinitesimal piece dy.

00:49 So dq by dy is the charge density, which is q divided by l, which means that dq equals q over l.

01:03 Now the electric field due to a charge de is dq divided by 4 pi epsilon nought r squared where r is the distance and then we've got this unit vector as well.

01:25 So r is the distance between the observation point and the place where the charge is and this unit vector points in the direction from the charge to the observation point.

01:41 So here, we've got a small positive charge here.

01:45 So the unit vector pointing away from it is this way.

01:48 So de is pointing away from the rod.

01:55 So now let's work out the total charge.

01:58 So e is going to be the integral from y equals zero to y equals l of dq over 4 pi epsilon naught and now r of y is going to be d plus y.

02:16 So we've got d plus y squared.

02:20 But we also have dq is q over l, dy.

02:24 So let's put that in here.

02:26 So q over l, dy.

02:35 Okay, so let's take out the constants we know.

02:40 So q over 4 pi epsilon naught.

02:43 And then the integral from nought to l, d y over d plus y squared.

02:51 And now we know what the integral of this is.

02:56 It's going to be minus 1 over d plus y, evaluated between nought and l.

03:07 Because when we take the derivative of this, we're going to get a minus sign from this being in the denominator, and then this is going to get squared.

03:14 So this is q over 4 pi epsilon nought.

03:18 Now let's evaluate this at l, so minus 1 over d plus l, and then we're going to take away the evaluation of 0, so plus 1 over d...

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