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Find the value of the z component of the total electric field

  • Thread starter blueyellow
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  • #1
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Homework Statement



Consider a generic point (-x,0,0). Calculate the distancrs of this point from all charges:
q at (-a,0,a)
q at (a,0,,a)
-q at (-a,0,-a)
-q at (a,0,-a)

and calculate the total electric field generated by each of them in terms of these distances, and the magnitude of the electric field produced by each charge. Find the z component of the total electric field

Homework Equations



E=(1/4pi*epsilon0)(q/r^2)*r-hat

The Attempt at a Solution



distance of q from charge -a,0,a:
=sqrt(2a^2 +x^2 -2ax)
distance from charge (-a,0,-a)
=sqrt(-2ax+x^2)
distance from charge (a,0,a)
=sqrt(2a^2 +x^2 +2ax)
distance from charge (-a,0,-a)
=sqrt(2a^2 +x^2 -2ax)

but I don't know how to use the equation to calculate the electric field because the r-hat term in the equation confuses me. please help
 

Answers and Replies

  • #2
gneill
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What do you think r-hat (that is, [itex]\widehat{r}[/itex] ) represents in the equation?
 
  • #3
241
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the unit vector? I still don't know how to deal with it though
 
  • #4
gneill
Mentor
20,797
2,776
Yes, they're unit vectors. They specify the directional information for the vector, while the Coulomb's Law bit provides the magnitude of the vector.

So, for each charge you need to find the magnitude and unit vector for its contribution to the field at the given point (-x, 0, 0). You'll probably want to fold the signs of the magnitudes into the unit vectors. Sum the individual vectors in the usual way to find the total field.
 
  • #5
241
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distance of q from charge -a,0,a:
=sqrt(2a^2 +x^2 -2ax)

magnitude of electric field=(1/4pi epsilon0)(2a^2 +x^2 -2ax)

value of z component:
=(1/4pi epsilon 0)(2a^2+x^2-2ax-(x-a)^2)

is this correct?
 
  • #6
gneill
Mentor
20,797
2,776
The magnitude should include the value of the charge which is causing the field. Note that the sign of the charge will affect things (how? What will that sign do to the direction of the electric field vector?).

It would probably be adequate for your purposes to leave the distance in the form
[tex] R = \sqrt{(a - x)^2 + a^2} [/tex]
Thus your magnitude for the field produced by the first charge, if the charge is +q, becomes
[tex] E_1 = \frac{k q}{R^2} = \frac{k q}{(a - x)^2 + a^2} [/tex]
with [itex]k = 1/(4 \pi \epsilon_o)[/itex]

The unit vector for the field direction is going to contain terms that come from the vector that represents the distance between the charge and the point. If r is the distance vector, R its magnitude, then the unit vector is r/|r| = r/R. Can you show the (components of the) distance vector and the unit vector, then extract the unit vector's z-component?
 
  • #7
241
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but what is r?
sorry I'm being stupid
 
  • #8
gneill
Mentor
20,797
2,776
but what is r?
sorry I'm being stupid
No worries :smile:

r is the distance vector between the charge (in this case the charge at (-a, 0, a) ) and the point (-x, 0, 0) . R is the magnitude of r.
 

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