Simple Electric Field Magnitude Question

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SUMMARY

The discussion focuses on calculating the electric field magnitude at point P due to two charged particles. The user initially miscalculated the magnitude of the electric field, stating it as \(\vec{E_{net}} = - \frac{2kq}{9} \hat{i}\). However, the correct expression is \(\vec{E_{net}} = -2 \left( \frac{kq\cos\theta}{25} \right) \hat{i}\), where \(\theta\) is defined as \(\arctan(4/3)\). The user recognized the need to incorporate the cosine of the angle into the final calculation for accuracy.

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jegues
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Homework Statement



See figure attached for problem statement.

Homework Equations





The Attempt at a Solution



I can see that the y-components of each particle will cancel each other out so we are solely looking at how the x-component forces of each particle affect the field at the point P.

These two forces will be added together because they are pointing in the same direction. These forces also happen to have the same magnitude.

So I see the resulting electric field to be as follows,

\vec{E_{net}} = - \frac{2kq}{9} \hat{i}

I keep getting the magnitude incorrect but I can see that I have the correct angle. (180 degrees)

What am I doing wrong for the magnitude?

EDIT: I found my mistake, I should have done,

\vec{E_{net}} = -2 \left( \frac{kqcos\theta}{25} \right) \hat{i}
 

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jegues said:
EDIT: I found my mistake, I should have done,

\vec{E_{net}} = -2 \left( \frac{kq\cos\theta}{25} \right) \hat{i}

How do you define θ ?

You should be able to determine cos(θ) and substitute that into your result.
 
SammyS said:
How do you define θ ?

You should be able to determine cos(θ) and substitute that into your result.

Where \theta = arctan(4/3)
 

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