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for the pentagon, there is a + charge at 1,  @2,  @3, + @4,  @5, and + @6. In part A, we are asled to find the net electric field at point P, which is half way between 1 and 2. Each side is 2 mm long, and each charge is an elemental charge. So, I did
E1=E2=kq/(r^2)...for r I used 0.001 m and got E1=E2=1.44x10^(3) N/C
E3=E5=kq/(r^2)...to figure out r, I used the law of cosines (using the 1mm side, 2mm side, and 108 degrees between them) and got the side to be 0.0025 m long so E3=E5=2.40x10^(4) N/C
E6=kq/(r^2)...to figure out r, I made a right triangle and used the 1 mm side and the 54 degree angle (from the hypotenuse line that bisects the 108 degree angle) I found the side to be 0.00137 m. so E6=7.67x10^(4)
E4=kq/(r^2)...to figure out r, I found the length from point 4 to point 6 by using a right triangle(the 1 mm side plus the 54 degree angle from the line that bisects the 108 degree angle) that length was 0.0017 m, and then you add that to the r from point 6 and you get r=0.00307 m
so E4=1.53x10^(4)
I drew an x coordinate parallel to the line connecting point 1 and 2 and I drew a y axis perpendicular to that line, and through point P.
so to find the x and y coordinates, I had to find the angles at which each of the electric field was at
For 1 and 2, the efield was at zero degrees, and for 4 and 6, the efield line was at 90 degrees. But, I had trouble with 3 and 5. For the angle for 3, could I used the law of sines and figure out the angle that the efield line(from 3) makes with the xaxis. if so, I got that to be 49.5 degrees, and so, at 3, the angle would be 310.5 degrees? and then for 5, from the law of sines, you get 49.5 degrees again, and so would that angle be 157.5 degrees?
For part B of the question, we are asked to find the net electric force at charge 6
F=(kq1q2)/r^2...since all of the distances from the outer charges to charge 6 are the same (0.0017m) all of the F's are the same=7.97x10^(23) N
when I had to find the x and y components of the net force, I drew a y axis straight down through charge 1, and an xaxis perpendicular to that and through charge 6. To find the angles, well since charge 1 is , the force would be directed downwards at 270 degrees. Now, I had trouble withthe rest of the charges. Since there are five charges, each makes a 72 degree angle with the center charge. So, for charge 2, it is directed away from the center charge, and would it be at 18 degrees? And in that same manner, charge 5 is directed away from the center charge, and would its angle be 162 degrees? For charge 4, it is directed towards the center charge, so would its angle actually be 54 degrees? And for charge 3, would its angle be 306 degrees?
For Part C, we are asked ro find the net electric field on point p (if it were an electron) (I used the same coordinate system as in part A)
To find each of the forces at each charge, I just multiplied the electric fields that i obtained by 1.6x10^(19) C. But then I had trouble again finding the angles in order to figure out the components. Since P is an electron, for charge 1, the electric force would be directed towards 1, and at 180 degrees. At charge 2, the electric force is directed towards the electron, again at 180 degrees. For charge 4 and charge 6, each of the electric forces are directed away from the electron, and therefore at 270 degrees. But , for charge 3 and 5, I wasn't sure. Both of the electric forces at directed towards the electron. FOr charge 5, since the angle we calculate before was 49.5 degrees, and charge 5 is in the first quadrant, would it be 49.5 degrees? And then, force charge 3, the angle was also 49.5 degrees that we calculated the first time, but the angle next to it was 40.5 degrees, and since the electric force for charge three will be in the third quadrant, will that angle be 90 degrees plus 40.5 degrees (130.5 degrees)?
One last thing....we were asked to find the net voltage at point P, which I did. Then we were asked to find the binding energy for the six charges. I did that also. But, I was wondering if the binding energy was supposed to equal the net voltage multiplied by the elemental charge. It doesn't work out that way, but I thought I remember hearing that.
Any help on anything would be so much appreciated. I know it is a lot....but thanks!
E1=E2=kq/(r^2)...for r I used 0.001 m and got E1=E2=1.44x10^(3) N/C
E3=E5=kq/(r^2)...to figure out r, I used the law of cosines (using the 1mm side, 2mm side, and 108 degrees between them) and got the side to be 0.0025 m long so E3=E5=2.40x10^(4) N/C
E6=kq/(r^2)...to figure out r, I made a right triangle and used the 1 mm side and the 54 degree angle (from the hypotenuse line that bisects the 108 degree angle) I found the side to be 0.00137 m. so E6=7.67x10^(4)
E4=kq/(r^2)...to figure out r, I found the length from point 4 to point 6 by using a right triangle(the 1 mm side plus the 54 degree angle from the line that bisects the 108 degree angle) that length was 0.0017 m, and then you add that to the r from point 6 and you get r=0.00307 m
so E4=1.53x10^(4)
I drew an x coordinate parallel to the line connecting point 1 and 2 and I drew a y axis perpendicular to that line, and through point P.
so to find the x and y coordinates, I had to find the angles at which each of the electric field was at
For 1 and 2, the efield was at zero degrees, and for 4 and 6, the efield line was at 90 degrees. But, I had trouble with 3 and 5. For the angle for 3, could I used the law of sines and figure out the angle that the efield line(from 3) makes with the xaxis. if so, I got that to be 49.5 degrees, and so, at 3, the angle would be 310.5 degrees? and then for 5, from the law of sines, you get 49.5 degrees again, and so would that angle be 157.5 degrees?
For part B of the question, we are asked to find the net electric force at charge 6
F=(kq1q2)/r^2...since all of the distances from the outer charges to charge 6 are the same (0.0017m) all of the F's are the same=7.97x10^(23) N
when I had to find the x and y components of the net force, I drew a y axis straight down through charge 1, and an xaxis perpendicular to that and through charge 6. To find the angles, well since charge 1 is , the force would be directed downwards at 270 degrees. Now, I had trouble withthe rest of the charges. Since there are five charges, each makes a 72 degree angle with the center charge. So, for charge 2, it is directed away from the center charge, and would it be at 18 degrees? And in that same manner, charge 5 is directed away from the center charge, and would its angle be 162 degrees? For charge 4, it is directed towards the center charge, so would its angle actually be 54 degrees? And for charge 3, would its angle be 306 degrees?
For Part C, we are asked ro find the net electric field on point p (if it were an electron) (I used the same coordinate system as in part A)
To find each of the forces at each charge, I just multiplied the electric fields that i obtained by 1.6x10^(19) C. But then I had trouble again finding the angles in order to figure out the components. Since P is an electron, for charge 1, the electric force would be directed towards 1, and at 180 degrees. At charge 2, the electric force is directed towards the electron, again at 180 degrees. For charge 4 and charge 6, each of the electric forces are directed away from the electron, and therefore at 270 degrees. But , for charge 3 and 5, I wasn't sure. Both of the electric forces at directed towards the electron. FOr charge 5, since the angle we calculate before was 49.5 degrees, and charge 5 is in the first quadrant, would it be 49.5 degrees? And then, force charge 3, the angle was also 49.5 degrees that we calculated the first time, but the angle next to it was 40.5 degrees, and since the electric force for charge three will be in the third quadrant, will that angle be 90 degrees plus 40.5 degrees (130.5 degrees)?
One last thing....we were asked to find the net voltage at point P, which I did. Then we were asked to find the binding energy for the six charges. I did that also. But, I was wondering if the binding energy was supposed to equal the net voltage multiplied by the elemental charge. It doesn't work out that way, but I thought I remember hearing that.
Any help on anything would be so much appreciated. I know it is a lot....but thanks!
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