A physics enthusiast decides to make mischief by creating an electric field at t

Question

A physics enthusiast decides to make mischief by creating an electric field at the center of the U.S. Pentagon. He puts electrostatic generators along the five corners of the pentagon, so that the net field at the center would be something tiny and nonzero and perhaps cause the body hair to stand up. If you are the U.S. General whose office is at the Pentagon center, what electric field will you notice in your office at origin? From the origin, the corners of the pentagon are at increments, all equidistant at some distance 1. You can assume the generators as point charges of q. Is this problem easier if you put one of the vertices on x-axis or y-axis? Indicate your choice. If you do not like working with unit vectors you may directly calculate the x and y components of Electric fields as well.
1. Use the unit vector notation to first find the direction vectors to each of the corners. Is the distance of each vertex from the origin necessary for this?
2. Calculate the magnitude of the electric field at the origin due to each vertex, and the net electric field at the origin using the previous step, in terms of the given variables. What is the direction and magnitude of the net field?
3. Could he have generated it with just one generator? Where would he need to place it?
4. Can he generate a field in some other direction? How would he do that?
5. What if the charges in three of the adjacent vertices are reversed?
express the Electric field in terms of the unit vectors.
6. If you so desire you may plug-in the numbers into the expressions calculated in each of the earlier parts-for l=20m and q=5 micrCoul.

Explanation / Answer

Each charge will create an E field at the center of the Pentagon; each with its own x/y compnent (Ex and Ey). All we have to do is find out all five Ex and Ey and then add them up to get the result of all the E components at the center.

Yes, it would be easier to have one of the axis on one of the vertex.

Each section of the pentagon has an angle of 72 deg (360/5). We will place the pentagon with its top vertex on top (on y axis) and base flat horizontally. The top vertex we'll call it A and going clockwise, B will be right vertex, C will be bottom right, D bottom left, and E on the left (picture this in ur mind!). Charge B will be at an angle 18 deg upward from the x-axis; samilarly for the charge E on the left side. C and D will both at 54 deg from the horizontal.

1. Yes, we need to know the distance from origin to vertex to get unit vector. We will get unit vectors:

A: It will only have vertical component, so it's Ey_A= -kqr/r^3 (negative for downard)

B: Horizontal- we dont need to find out cauz it will cancel out with Ex at E.

Vertical- Ex_B= -kqr/r^3 sin18

C: Hoizaontal- we dont need to find out cauz it will cancel out with Ex at D.

Vertical- Ey_C= kqr/r^3 sin54

D: horizontal: don't need to find out

Vertical- Ey_D = same as Ey_C = kqr/r^3 sin54

E: hoizontal- dont need to find out

vertical- Ex_E = same as Ex_B = Ex_B= -kqr/r^3 sin18

2. we would add all the E components together. There is only vertical comonent; all horiz cancels out. We also use E, not unit vectors.

E= Ey_A + Ey_B + Ey_C + Ey_D + Ey_E

= -kq/r^2 -kq/r^2 sin18 -kq/r^2 sin18 + kq/r^2 sin54 +kq/r^2 sin54 (simplify this. answer will be neg meaning downard. let's called this [2])

3. yes, solve for E field one charge created by one charge somewhere on the y axis. Solve for r

[2]=kq/r^2 --> r= ( kq/[2] )^.5

4. no, it is direction can't be changed because the location of charges can't be changed.

5. Let's let A B C become -q charged. This would make all the vertical component into the same downward direction.

E' = kq/r^2 + kq/r^2 sin18 + kq/r^2 sin18 + kq/r^2 sin54 +kq/r^2 sin54

6. solve this yourself; just plug it in. r=l=20m and q=5microC

:D

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