<p>Hey.. I had a few other questions which I'd appreciate some help with:</p>
<p>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.</p>
<p>I essentially found the force on one corner ( by using x and y components )</p>
<p>Hence, for X = 2.29E7 + 0 + 3.24E7
For Y = -2.29E7 +3.24E7</p>
<p>I found the force to be 5.6E7. is this correct?</p>
<hr>
<p>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?</p>
<p>not sure how to go about this. Do I say:</p>
<p>Q1+Q2+Q3=0
Hence, F/E+F/E+F/E=0 ????</p>
<p>Thanks much.</p>
<p>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. </p>
<p>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</p>
<p>The second one
The field created by the first one is inward, and the positive one is outward.
This is just asking about electric fields.
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.</p>
<p>
[quote]
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.</p>
<p>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
[/quote]
</p>
<p>kinda lost on the second paragraph.. I measured 45 degrees but not sure where you got the Sqrt(2) from... </p>
<p>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.</p>
<p>
[quote]
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.
[/quote]
</p>
<p>Wow? its that simple? </p>
<p>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?</p>
<p>THanks</p>
<p>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.</p>
<p>(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.</p>
<p>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.</p>