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# Physics help

edited February 2006
Hey.. I had a few other questions which I'd appreciate some help with:

Q3) A charge of 6 mC is placed at each corner of a square .100 m on a side. Determine the magnitude and direction of the force on each charge.

I essentially found the force on one corner ( by using x and y components )

Hence, for X = 2.29E7 + 0 + 3.24E7
For Y = -2.29E7 +3.24E7

I found the force to be 5.6E7. is this correct?

__________________________________________________ __________

Q4) A +4.75 micro C and a -3.55 micro C charge are placed 18.5 cm apart. Where cana third charge be placed so that it experiences no net force?

Q1+Q2+Q3=0
Hence, F/E+F/E+F/E=0 ????

Thanks much.
Post edited by Unknown User on

## Replies to: Physics help

• Registered User Posts: 660 Member
For the top right one, you should have KQQ/r^2 pushing right, and then that is also pushing up, and then for the far one, its up and right and its 1/2kQQ/r^2 and then its 45degree angles and thats the hypotenuse, so divide that by root 2 and add it to pushing right and pushing up.

Each charge is diagonally outward, well the net vector is and it is KQQ/2(r^2) (one opposite corner) + sqrt(kQQ/r^2 * 2) (its 2 because one is to the right, and one is up) so its 1/2X + root2X and X just KQQ/.1^2 where K = 9*10^9

The second one
The field created by the first one is inward, and the positive one is outward.
If possible spot <-pos-> >-neg-< (possible spot) and you can eliminate the K its just an extra calculation
so the pos 4.75/(18.5cm + X)^2 + -3.55/(x)^2 = 0 (no force)
Using an 89, the values of X are 118cm and -8.577 meaning a spot there are 2 spots
- 8.57cm + 118cm+18.5cm all distances relative to the neg charge.
• Administrator Posts: 32,363 Senior Member
For the top right one, you should have KQQ/r^2 pushing right, and then that is also pushing up, and then for the far one, its up and right and its 1/2kQQ/r^2 and then its 45degree angles and thats the hypotenuse, so divide that by root 2 and add it to pushing right and pushing up.

Each charge is diagonally outward, well the net vector is and it is KQQ/2(r^2) (one opposite corner) + sqrt(kQQ/r^2 * 2) (its 2 because one is to the right, and one is up) so its 1/2X + root2X and X just KQQ/.1^2 where K = 9*10^9

kinda lost on the second paragraph.. I measured 45 degrees but not sure where you got the Sqrt(2) from...

THere are four charges q1,q2,q3,q4. Q1 is on top left and im trying to measure force on that charge. Hence, I essentially thought that since its a square, one of the charge ( bottom left ) acting on Q1 will be in the x dimension only, the other charge will be only in y dimension ( the one on the top right ) and the third charge ( diagonally across and bottom right) will have both x and y component.

so the pos 4.75/(18.5cm + X)^2 + -3.55/(x)^2 = 0 (no force)
Using an 89, the values of X are 118cm and -8.577 meaning a spot there are 2 spots
- 8.57cm + 118cm+18.5cm all distances relative to the neg charge.

Wow? its that simple?

So that means the charges can be in three positions , 8.57 am away, 118 cm away and 18.5 cm away from the negative charge?

THanks
• Registered User Posts: 774 Member
Wazzup:
(1) In a right-angled triangle with angles of 45,45 and 90 degrees, the sides are 1,1, and (hypotenuse) sqrt(2) . So a diagonal force of F units along a 45% incline can be broken up into equal x- and y- components of F/sqrt(2) each.

(2) I think there are just 2 solutions, not 3. The two feasible values of x (measured relative to the position of the negative charge) are given by the solutions to the implied quadratic equation.
• Registered User Posts: 660 Member
the 18.5 is just the distance they are apart, it isnt a location of zero potential. Thats what you are essentially finding, positions of zero electric potential.
This discussion has been closed.