What is the Zero Field Point Between Two Charged Particles on the x-axis?

[Abelard]

Homework Statement

Two particles are situated on the x axis. The particle q1, with 34 excess electrons, is situated at the point x = 100 μm. The other particle q2, with 17 excess electrons, is located at the origin. Give the x value of a point between the particles where the strength of the field they generate is zero.

Homework Equations

So kQ/r^2 is important here. The addition of electric field strength must give 0 N/C.

The Attempt at a Solution

I'm not even sure where that x value would fall into. Would it be between the two charges or to the right of -34e or to the left of -17e. kq1/x^2 +kq2/(x-100e-6)^2=0 is the equation I tried to solve for x, but the function is asymptotic to the x axis, so, the answer is infinity. But that wasn't the answer in the back of the book, so figure it out.

 

[Abelard]

Or I should've added x+100e-6 rather than subtracting it.

 

[gneill]

Pick an arbitrary point on the three regions of the line through the charges (to the left of, between, to the right of).

Without doing any numerics, sketch estimated field vectors for the contributions of both charges. For example, to the right of both charges, both field vectors will point in the same direction, to the left.

Determine the region where the two vectors have a chance of summing to zero, and concentrate your efforts there.

 

[Abelard]

It will be between the charges. kq1/x^2 = -kq2/(x-100e-6)^2; where this holds true would be the x. But the calculator can't find the intersection.

 

[gneill]

It will be between the charges. kq1/x^2 = -kq2/(x-100e-6)^2; where this holds true would be the x. But the calculator can't find the intersection.

Suppose you were to collect all your forces on one side of the equality and set to zero:

kq1/x^2 + kq2/(x-100e-6)^2 = 0

Both q1 and q2 have the same sign (they're both negative); Your expression can never be zero. One or the other of the terms needs to have its sign reversed.

 

[Abelard]

OK, so kq1/x^2 = kq2/(100e-6-x)^2 would be a right equation. Now I got it. x=4.14e-5m. Thanks a lot.