Thursday, December 17, 2009

Could someone help my son and I figure out this physics problem?

Find the electric field and the potential at the center of a square of side 2.0cm with a charge of +9.0 microC at one corner of the square and with charges of -2.0 microCoulombs at the remaining 3 corners.





Any help I would greatly appreciate, check my account all the time so best answer will get 10 points as soon as possibleCould someone help my son and I figure out this physics problem?
let E= Electric field


let q= charge strength


let r= distance from charges.


k is a physics constant= 8.99 X 10^9


E=kq/r^2


So whenever you do a physics problem it is always good to draw a picture and label and draw in all the relevant forces.


the two -2 micro coloumb charges at opposite sides of the square will have the same E-field at the center of the square but in opposite directions, since the distance to the center and charge is constant, the E-field due to these 2 charges alone will be 0, so we can ignore them.


since electric field always goes from positive charges to negative charges, the +9 micro coloumb charge will produce an E-Field in the direction of the -2 micro coloumb charge on the opposite side of the square. Also, the -2 micro coloumb charge will produce an E-field towards itself, enhancing the E-field of the +9 micro coloumb charge.


to solve for r, note that if u draw a diagonal throigh the square it forms two 45-45-90 triangles whose side ratios are 1:1:sqrt(2). in this case 2:2:2sqrt(2). but we only want half this distance since we want the distance to the center of the square so, r=2sqrt(2)/2=sqrt(2) *10^-2 m


plugging in:


kq/r^2 - kq/r^2=k*(9*10^-6)/(sqrt(2)*10^-2)^2 - k*(-2*10^-6)/(sqrt(2)*10^-2)^2 =11k*10^-6/(2*10^-4)= 49 4450000 N/C





electric potential is a scalar value which means that it does not have a direction like E-field does.


let v= electric potential


v=kq/r


we must add all 4 potentials produced from each of the 4 spheres.


kq/r+kq/r+kq/r+kq/r


=k/r(-2+-2+-2+9)*10^-6


=k/r(3*10^-6)


=8.99*10^9*3*10^-6/(sqrt(2)*10^-2)


=1 907 066.99 VoltsCould someone help my son and I figure out this physics problem?
Well, I am guessing that this is supposed to be in free space, meaning permittivity is that of free space, or 8.854x10^(-12) F/m.





E = (Q1(r-r1))/(4*Pi*(8.854*10^(-12))*(|r-r1鈥?br>




Sum this for all four charges and you will find the electric field. The easiest way to do it is to let r = 0ax + 0ay. This just means that the center of the square is at the origin.





Therefore, E1 = (9*10^(-6))*(0-(1x10^(-2)ax + 1x10^(-2)ay))/(4*Pi*(8.854*10^(-12))*(|0鈥?+ 1x10^(-2)ay)|^3))





E1= -285988ax + -285988ay


Repeat for the other sides and sum them all up.


This yields a final answer of E = -349541.1494 ax + -34954111494 ay V/m.


This is assuming the charge of +9microC is at the top right. If it is moved, change the signs accordingly. If you need the magnitude only of E, it is 494325.834 V/m.





For potential, V = - integral(E x dl) . This means take each E you calculated earlier, and integrate it along the line between the center and the corner. Then sum up the total V. This means for E1, V1 = integral ( - Eax dx from 0 to 1x10^(-2) - Eay dy from 0 to 1x10^(-2)).











Edit: Note, k, the 'physics constant' in the other answer is supposed to be 8.854 x 10^(-12). 8.9x10^(-12) is close enough, but the exponent on ten is definitely -12, according to an Electromagnetics textbook I am looking at right now.

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