Zero Point of field between two charges

So E1=E2. So get y in terms of the charges, and then use your equations to find the actual value of y.In summary, the problem involves two point charges located along the y-axis. The total electric field at a point between them (other than infinity) is zero. To solve this, the signs of the E-field components must be taken into account. Equating the electric fields from each charge and solving for y will give the location at which the total electric field is zero.
  • #1
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Homework Statement


Two point charges lie along the y-axis. A charge of [itex]q_1 = -9.0 \times 10^{-6}\;C[/itex] is at [itex]y = 6.0m[/itex], and a charge of [itex]q_2 = -8.0 \times 10^{-6}\;C[/itex] is at [itex]y = -4.0m[/itex]. Locate the point (other than infinity) at which the total electric field is zero.


Homework Equations


[tex]E = \frac{k|Q|}{d^2}[/tex]


The Attempt at a Solution


All I know is that it's between the two points. Basically, I don't know how to set this problem (or any like it) up. I'm not asking for the answer, just a nudge in the right direction.
 
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  • #2
The E field at any point is a vector.

In 1D (i.e. along the y-axis in your Q) the E field at a given point is either +ve or -ve, so you have to get the signs right. (Your eqn. won't work as is. Remove the magnitude sign.)

Then add the E-field components to find E as a function of y. Diff to find point where dE/dy=0
 
  • #3
Not sure why it was mentioned to take a derivative, but the rest is okay. (the potential isn't mentioned here)
 
  • #4
robb_ said:
Not sure why it was mentioned to take a derivative, but the rest is okay. (the potential isn't mentioned here)

Sorry- you're right. You need to equate the fields to find when they're equal and opposite.
 

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