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

In summary: The z-component of the unit vector is just the length of the vector multiplied by the z-component of the charge.
  • #1
blueyellow

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
 
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  • #2
What do you think r-hat (that is, [itex]\widehat{r}[/itex] ) represents in the equation?
 
  • #3
the unit vector? I still don't know how to deal with it though
 
  • #4
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
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
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
but what is r?
sorry I'm being stupid
 
  • #8
blueyellow said:
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.
 

1. What is the equation for finding the value of the z component of the total electric field?

The equation for finding the value of the z component of the total electric field is E_z = E * cos(theta), where E is the magnitude of the total electric field and theta is the angle between the total electric field and the z-axis.

2. How is the z component of the total electric field related to the x and y components?

The z component of the total electric field is related to the x and y components through the Pythagorean theorem. The magnitude of the total electric field, E, can be found using the equation E = sqrt(E_x^2 + E_y^2 + E_z^2).

3. What is the unit of measurement for the z component of the total electric field?

The unit of measurement for the z component of the total electric field is typically newtons per coulomb (N/C). This unit represents the force per unit charge experienced by a charged particle in the z-direction due to the electric field.

4. Can the z component of the total electric field be negative?

Yes, the z component of the total electric field can be negative. This indicates that the electric field points in the negative z-direction, opposite to the direction of the z-axis.

5. How can the value of the z component of the total electric field be used in practical applications?

The value of the z component of the total electric field can be used to calculate the force on a charged particle in the z-direction, or to determine the direction of motion of a charged particle in an electric field. It is also important in understanding the behavior of electric fields in three-dimensional space.

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