Troubleshooting Electric Field Calculation: Qs, Rs, and N/C

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SUMMARY

The forum discussion centers on calculating the electric field at a midpoint between two charges, specifically a -8.0 µC and a +6.0 µC charge, located 4 cm apart. The initial calculation incorrectly omitted squaring the distance for one of the charges, leading to an erroneous result of 6,300,000 N/C. The correct approach involves applying Coulomb's law and the superposition principle to compute the electric fields due to each charge separately and then summing them to find the net electric field.

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joej
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I'm doing the following problem, it seems to me that I am doing all the correct steps but my answer differs from the answer key, am I missing something?

1. What is the magnitude and the direction of the electric field at a point midway in between a -8.0uC and a +6.0uC change 4cm apart.


This is what I'm doing:

E = ( k * (Q1 / r1^2) ) * ( 1 + ( (Q2 / Q1) / (r2^2 / r1^2) ) )

==

(9x10^9 * ( (8 * 10^-6) / ( 2 * 10^-2) ) ) * ( 1 + ( (6 / 8) / (2 / 2) ) )

==

3600000 * 1.75 = 6300000 N/C


now... that is what I'm getting, could somoen please point out to me what I am forgetting to do, if anything.



__________edit

stupidity strikes again... forgot to square r1
 
Last edited:
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Why write E in such a strange way? (to me anyway)

Use Coulomb's law and the superposition principle.
Calculate the field due to Q1 and the field due to Q2, then add.
 
Hello joej...

The formula for electric field at a point distant r units from a charge Q is:

[tex]\vec{E} = \frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^2}\hat{e}_{r}[/tex]

The direction of the field is radial. In your problem you need to compute

[tex]\vec{E}_{net} = \vec{E_{1}} + \vec{E_{2}}[/tex]

the two terms arising due to the electric fields produced (independently--make a note of this, the superposition principle) has been used here) by the two charges. Have you used this fact? :confused:

Cheers
Vivek
 

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