Calculating Electric Field Using Coulomb's Law

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Jrlinton
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Homework Statement


electri field.PNG


Homework Equations



Ex=1/k*1/(a^2/2)*1/(sqrt2)*cos(theta)*(sum of charges)
Ey=1/k*1/(a^2/2)*1/(sqrt2)*sin(theta)*(sum of charges)

The Attempt at a Solution


So first off I can see that I don't need to calculate the force in the x direction as each q cancels out when arranged appropriately
For Ey I get:
Ey=1/k*1/(.067^2*2)*1/sqrt2*sin(45degrees)*(-6.31nC+18.6nC+18.6nC-6.31nC)*10^-9
=1.523E-16 N/C

Use Pythagorean theorem but that is unnecessary do to an absence of an x component
 
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just go "by parts" try finding y comp of charge 1 and 4 then 2 and 3 and just add them up. or realize that charges 1 and 2 resultant y component is the same as 3 and 4 so you just have to do one calculation and multiply by 2
 
I did as you said using parts and still came up with the same answer of 1.523E-16 N/C.
1/k*1/(0.67^2*2)*1/sprt2*sin(45deg)*(-631nC+18.6nC)*10^-9
=7.61E-17
*2
=1.523E-16
 
Jrlinton said:
I did as you said using parts and still came up with the same answer of 1.523E-16 N/C.
1/k*1/(0.67^2*2)*1/sprt2*sin(45deg)*(-631nC+18.6nC)*10^-9
=7.61E-17
*2
=1.523E-16
that doesn't seem right... what are you using to calculate the field? E=F/qtest this q is normally a test charge (1C) and F = kQ*qtest/r2 so the test charge cancels out which makes sense since you want the field created by the source charge.
so Fx=Fcos(theta)
ps: r is the distance between the charge and the point where you are evaluating the field
ps2: yeah it doesn't seem to be a unit problem, i don't know where you got the 1/k or the sqrt2

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefor.html#c1
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elefie.html
maybe those will help clearing things up
 
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