Find the electric field at the point given

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Homework Help Overview

The problem involves calculating the electric field at a specific point (x) due to two point charges: +50μC and -10μC, positioned at known distances from point x. The context is within electrostatics, specifically focusing on electric fields generated by point charges.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using the formula for electric field intensity, E=kq/r^2, and question which charge to use for 'q'. There is also mention of vector addition of electric fields from both charges.

Discussion Status

The discussion is ongoing, with participants exploring different aspects of the problem, including the correct application of the electric field formula and the distances involved. Some guidance has been offered regarding the use of vector addition and the interpretation of 'r' in the context of the electric field equations.

Contextual Notes

There appears to be some confusion regarding the appropriate distances to use for calculating the electric field from each charge, as well as the overall approach to combining the electric fields from multiple sources.

xxabr
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Homework Statement



Find the electric field at the point given, x.

(+50uC)-------0.45m---- (-10uC) -----0.30m----- ( x )


Homework Equations


E= kq/r^2

Maybe, E= Fe/q and Fe= kq1q2/r^2


The Attempt at a Solution



q1= 50x10^-6
q2= -10x10^-6
q3= ?

I have no idea what equation to start with or where to start.
If I were to use E=kq/r^2
I don't know which charge to use for 'q'
Please help, thanks.
 
Last edited:
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Use vector addition of Electric field intensity.
The electric Field at x will be
Ex = E1 + E2
where E1 and E2 are the electric field intensity vectors due to 50 and -10uc resp.
 
Alright so,
If I use E=kq/r^2 for 50x10^-6C ... R would be 0.45m
and for -10x10^-6C ... R would be 0.30m?
 
xxabr said:
Alright so,
If I use E=kq/r^2 for 50x10^-6C ... R would be 0.45m
and for -10x10^-6C ... R would be 0.30m?

No and Yes.
Check your book where you derived electric field intensity due to a point charge. Find out what was 'r' there
 

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