Question 1 -
Coulomb's law for a line of charge uniformly distributed is
![1731_figure.png](https://secure.tutorsglobe.com/CMSImages/1731_figure.png)
a. We are interested in the electric field at point A from a semicircle of charge with charge density ρl. You should NOT solve this integral, only set it up to be solved.
What would you substitute for dl, r-, and r-' ? What are the limits of the integral? Remember that r-' is a vector that points from the origin to some piece of charge, and r-' is a vector that points to where you want to calculate the electric field. The idea of the integral is to sum up the electric field from every little piece of charge.
Use cylindrical coordinates exclusively, and set the origin at point A. Let φ = 0 be the direction upward on the page, away from the semicircle.
Question 2 -
Have you ever driven by a wind energy farm? If so, you've probably noticed that the turbines are hoisted above the ground. This is because the wind is in general higher as you go up in altitude from the ground. If the wind speed were constant within a region, the power collected from a wind turbine is fairly simple and can be written as P = ½CρAv3 where C is an efficiency coefficient (a constant), ρ is the air density (kg/m3), A is the area swept/subtended by the rotor blades, and v is the wind velocity.
But in reality, the wind speed varies with altitude, so that above equation for power isn't fully correct. Let's model the wind speed as v(Z) = (vxo + az)x^ m/s, where z^ points vertically upward, x^ is along the ground and in the direction of the wind, y^ is along the plane of the turbine. vxo is the wind speed on the ground and a is a constant defining how the wind speed increases with height. See below diagram.
![449_figure1.png](https://secure.tutorsglobe.com/CMSImages/449_figure1.png)
a. Set up a surface integral to calculate the wind power collected by the turbine. I suggest that you use Cartesian coordinates, but justify this to yourself. The difficult part will be determining the limits of the double-integral. You do not need to solve or simplify the integral, just leave it as a surface double-integral, with the variables substituted. Assume the wind turbine has radius r.
b. Now let's put some numbers to this. A particularly powerful and large wind turbine, the Vestas V-164, has a radius of 82 m (or diameter 164 m), and it can generate a maximum of 8 MW. Assume the turbine is located barely above the ground, meaning the bottom of the turbine's circle just barely goes over the surface. Assume that a = 0.1 m/s/m, meaning that the wind speed increases by 0.1 m/s for each meter of altitude increase. Assume the wind speed is zero on the ground, in other words vx0 = 0. Assume that C = 0.4. The density of air is 1.225 kg/m3. Now solve your surface integral in part (a). How much power does the wind turbine generate? Note: To help with the integration step, I'll be posting a quick refresher (and a quick numerical shortcut you may not seen before) to do integration by parts.
c. If the turbine is rotated by 45 degrees (meaning it faces northwest instead of west, for example), what happens to the generated power? Justify your answer using vector math.
Provide detail answers.