Question

Given two point charges q1 and q2 = 4q1, find the position of a third charge q2 relative o the other charges, such that the resultant force on q3 is zero.

Answer 1

I dont really get it how to draw this up..

Answer 2

shouldn't that be "a third charge q3" not "a third charge q2?"

Answer 3

yes... damn it..

Answer 4

lol so we have two charges

Answer 5

if we place a third charge in the middle, the two forces from q1 and q2 will act in opposite directionsthere is some point where the force on q3 will be zero

Answer 6

Fq1-Fq2=0

Fq1=Fq2

Answer 7

yes, now we need a way to represent Fq1 and Fq2, which means we need to set up a coordinate system to represent the radius between q1 and q2

Answer 8

?

Answer 9

perfect

Answer 10

do you know what to do next?

Answer 11

hmm maybe since Fq1=Fq2

\[k*q_1*q_3/x^2=k*q_2*q_3/(r-x)^2\]

Answer 12

then solve for r?

Answer 13

This reminds me of a gravitational assignment I had on that chapter..

Answer 14

yep
except you aren't solving for r, you are solving for the variable that represents the distance of q3 from q1

Answer 15

yeah you can do most of the same calculations with gravity and static charges in classical mechanics

Answer 16

I have a difficulty with the mathematics here to make the equation right.. But I want it to say r-x=?

Answer 17

you mean you can't do the algebra?
first substitute for q1 and q2, then see what cancels

Answer 18

\[k\frac{q_1q_3}{x^2}=k\frac{q_2q_3}{(r-x^2)}\]does anything cancel?

Answer 19

yeah but q3 and k must cancel?

Answer 20

yep

Answer 21

\[\frac{ q_1 }{ x^2 }=\frac{ q_2 }{ (r-x)^2 }\]

Answer 22

right, now substitute the values for q1 and q2

Answer 23

plug in the values*

Answer 24

Do I have value for q1?

\[\frac{ q_1 }{ x^2}=\frac{ 4q_1 }{ (r-x)^2 }\]

\[\frac{ (r-x)^2 }{ x^2 } = 4\]

\[(r-x)^2=4x^2\]

Answer 25

\[r-x=\sqrt{4x^2}\]

Answer 26

no, but it doesn't matter, what you have is correct

Answer 27

\[r-x=\pm2x\]

\[r=\pm3x\]

Answer 28

?

Answer 29

wait I must solve for x?

Answer 30

well first of all, if you were going to consider the negative root it would appear here:\[r-x=\pm\sqrt{4x}\]second, if we follow that through it would be\[r-x=\pm2x\\r=\{-x,3x\}\]however we can ignore the negative root completely because it is on the left of q1, and there is no way for the forces from q1 and q2 to cancel there because they act in the same direction.

Answer 31

should be \(x^2\) on the first line under the root...

Answer 32

and yes, you must solve for x

Answer 33

so take it again from here\[r-x=\sqrt{4x^2}\]and only consider the positive root

Answer 34

x=r/3?

Answer 35

yes :)

Answer 36

thanks alot :D

Answer 37

welcome!