Let l = Q/L = linear charge density. The semi-circle has a length L which is half the circumference of the circle. So w can relate the radius of the circle to L by
<span>C = 2L = 2*pi*R ---> R = L/pi </span>
<span>Now define the center of the semi-circle as the origin of coordinates and define a as the angle between R and the x-axis. </span>
<span>we can define a small charge dq as </span>
<span>dq = l*ds = l*R*da </span>
<span>So the electric field can be written as: </span>
<span>dE =kdq*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>dE = k*I*R*da*(cos(a)/R^2 I_hat + sin(a)/R^2 j_hat) </span>
<span>E = k*I*(sin(a)/R I_hat - cos(a)/R^2 j_hat) </span>
<span>E = pi*k*Q/L(sin(a)/L I_hat - cos(a)/L j_hat)</span>
Oil <span>must be mined from underground
Oil is NOT a renewable resource
Oil does NOT release more toxins than coal
Oil also is NOT easily replenished
</span>
There are no appropriate units for power on the list you provided
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Question 7:
</h2>
![\huge\text{Graphs:}](https://tex.z-dn.net/?f=%5Chuge%5Ctext%7BGraphs%3A%7D)
The graph of
• The I-V for Ohmic Metal wire conductor at constant temperature always shows a straight line between the Current(I) plotted at Y axis and Voltage(V) plotted at X axis. Picture 1
• The I-V graph for Diode shows that first the current is zero but as we increase the potential difference(voltage), it results in the increase in the current. Picture 2
<h2>_____________________________________
</h2><h2>Question 8:
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A diode is a device that allows current to flow in only one direction.
Forward Bias, When a diode is forward bias (a voltage in the "forward" direction) then the P-side of the diode is attached to the positive terminal and N-side is fixed to the negative side of the battery which is connected, current flows freely through the device. The forward bias decreases the thickness of potential barrier(The potential barrier barrier in which the charge requires additional force for crossing the region)
Reverse Bias, When a diode is Reverse bias(a voltage in the "backward direction) then the P-side of the diode is connected to the negative terminal and N-side is connected to the positive terminal of the battery which is connected. The reverse bias increases the thickness of the potential barrier resulting in the flow of no current.
![\Large\textbf{Answer to the Question "Resistance"}](https://tex.z-dn.net/?f=%5CLarge%5Ctextbf%7BAnswer%20to%20the%20Question%20%22Resistance%22%7D)
The Forward bias decreases the resistance of the diode whereas the reversed bias increases the resistance of the diode. As in forward biasing the current is easily flowing through the circuit whereas reverse bias does not allow the current to flow through it.
<h2>_____________________________________
</h2><h2>Best Regards,
</h2><h2>'Borz'
</h2>