Zero Point of field between two charges

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To find the point where the total electric field is zero between two negative point charges, it's essential to recognize that the electric field is a vector quantity and can be either positive or negative depending on the position along the y-axis. The solution involves calculating the electric field due to each charge and setting their magnitudes equal to find the balance point. The equation for the electric field should not include absolute values, as the direction (sign) is crucial. Differentiating the electric field function can help identify the point where the fields cancel each other out. Ultimately, the goal is to locate the position along the y-axis where the electric fields from both charges are equal and opposite.
duelle
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Homework Statement


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

Homework Equations


E = \frac{k|Q|}{d^2}

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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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
 
Not sure why it was mentioned to take a derivative, but the rest is okay. (the potential isn't mentioned here)
 
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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