# Electric forces and electric fields

• sportsrules
In summary: E1=14.4x10^(-24) JE2=6.4x10^(-27) JE3=3.6x10^(-28) JE4=1.2x10^(-31) JE5=5.2x10^(-34) JE6=3.4x10^(-36) J
sportsrules
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 e-field was at zero degrees, and for 4 and 6, the e-field 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 e-field line(from 3) makes with the x-axis. 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 x-axis 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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Well first of all, pentagon or hexagon? Your sentence "for the pentagon, there is a + charge at 1, - @2, - @3, + @4, - @5, and + @6" is mildly confusing.

sportsrules said:
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-

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 e-field was at zero degrees, and for 4 and 6, the e-field 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 e-field line(from 3) makes with the x-axis. 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?

Any help on anything would be so much appreciated. I know it is a lot...but thanks!

All of it is a bit too much for someone to work out on their own. It looks to me like you have already found the distance P3 from law of cosines knowing angle 2 and sides P2 and 2-3. The law of sines will tell you angle 2P3, and you can extend line 3P to get the same angle in the second quadrant of your coordinate system (assuming y is directed out of the pentagon. If angle 2P3 is the 49.5 degrees (looks about right), then you seem to be on track.

## 1. What is an electric force?

An electric force is a fundamental force of nature that exists between two charged particles. It can either attract or repel the particles depending on their charges. Like charges repel each other, while opposite charges attract each other.

## 2. How is the strength of an electric force determined?

The strength of an electric force is determined by the magnitude of the charges involved and the distance between them. The larger the charges and the smaller the distance between them, the stronger the electric force will be.

## 3. What is an electric field?

An electric field is a region of space around a charged particle or collection of charged particles where an electric force can be felt. It is created by the presence of electric charges and can exert a force on other charged particles placed within the field.

## 4. How is the direction of an electric field determined?

The direction of an electric field is determined by the direction that a positive test charge would move if placed in the field. It is always directed away from positive charges and towards negative charges.

## 5. What is the relationship between electric force and electric field?

Electric force and electric field are closely related. The electric field is a way of describing the influence that a charged particle has on the space around it. The electric force is the actual interaction between two charged particles within that electric field.

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