Find the x component of the Electric Field

In summary, the problem asked to find the general formula for the x component of the electric field, given that the charge density only varies with x throughout all space. Using the Poisson equation, it was determined that the x component of the electric field is equal to the charge density divided by the permittivity of free space. To solve for the constant of integration, boundary conditions were needed and it was assumed that the electric field at x=0 is equal to 0. From this, the constant of integration was determined to be equal to the negative of the charge density at x=0 divided by the permittivity of free space.
  • #1
themagiciant95
57
5

Homework Statement


Find the general formula for the x component of the electric
field if the charge density p varies only with x throughout all
space.

Homework Equations

The Attempt at a Solution



I started using the poisson equation [tex]\bigtriangledown \bullet \bar{E} = \frac{p}{\varepsilon_{0}}[/tex]
From the problem text, I know that p has only the x component and so also E has only the x component.
With these information, i tried to calculate the poisson equation, obtaining :

[tex]\frac{\partial E_{x}}{\partial x} =\frac{p(x)}{\varepsilon_{0}}[/tex]

But, i don't know how to continue the calculations... Can you help me ?
 
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  • #2
Since the variation is only in x, the left hand side is [itex] \frac{dE_x}{dx}[/itex], not [itex] \frac{\partial E_x}{\partial x}[/itex]. Now multiply through by dx and integrate both sides.
 
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  • #3
phyzguy said:
Since the variation is only in x, the left hand side is [itex] \frac{dE_x}{dx}[/itex], not [itex] \frac{\partial E_x}{\partial x}[/itex]. Now multiply through by dx and integrate both sides.

Do i have to integrate it as an definite or indefinite integral ? In the latter case, how can i manage the constant of integration ? Thanks
 
  • #4
Well, since you aren't told the boundary conditions, you will have to make some assumptions. You could assume for example that Ex at -∞ is zero. Or you could include the value of Ex at some point in your calculation.
 
  • #5
For example, if i include the value of Ex in a point (it's sufficient ?), how can i use this value to calculate the constant of integration ? Thanks again
 
  • #6
Why don't you show us the calculation with the constant of integration included? Then if it isn't clear how to deal with it we can make suggestions.
 
  • #7
phyzguy said:
Why don't you show us the calculation with the constant of integration included? Then if it isn't clear how to deal with it we can make suggestions.

[tex]
\int dE_{x} =\frac{1}{\varepsilon _{0}}\int p(x)dx
[/tex]

[tex]E_{x}(x)=\frac{1}{\varepsilon _{0}}P(x) + c[/tex]

As boundary i chose:

[tex]x=0 \rightarrow E_{x}=0[/tex]
so
[tex]c=-\frac{1}{\epsilon _{0}}P(0)[/tex]

is this correct ? Thanks
 
  • #8
up
 
  • #9
It looks OK to me. Without knowing more about the boundary conditions, I think something like this is the best you can do. You could try asking your teacher for more clarification.
 
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  • #10
Thanks so much for your help :)
 

Related to Find the x component of the Electric Field

What is the equation for finding the x component of the Electric Field?

The equation for finding the x component of the Electric Field is: Ex = (k * Qx) / r^2, where Ex is the x component of the Electric Field, k is the Coulomb's constant (8.99 x 10^9 N * m^2 / C^2), Qx is the charge of the source of the field in the x direction, and r is the distance between the source and the point of interest.

How do I determine the direction of the x component of the Electric Field?

The direction of the x component of the Electric Field is determined by the direction of the charge Qx. If Qx is positive, the x component of the Electric Field points in the same direction as the x-axis. If Qx is negative, the x component of the Electric Field points in the opposite direction of the x-axis.

Can the x component of the Electric Field be negative?

Yes, the x component of the Electric Field can be negative. This indicates that the direction of the Electric Field is opposite to the direction of the x-axis, which happens when the source charge Qx is negative.

How is the x component of the Electric Field affected by changes in distance?

The x component of the Electric Field varies inversely with the square of the distance between the source charge and the point of interest. This means that as the distance between the two increases, the magnitude of the x component of the Electric Field decreases.

What is the unit of measurement for the x component of the Electric Field?

The unit of measurement for the x component of the Electric Field is Newtons per Coulomb (N/C). This unit represents the force per unit charge exerted by the Electric Field in the x direction.

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