How Does a Uniform Electric Field Affect Charge Density Distribution?

[smantics]

Homework Statement

A space has an uniform electric field
E=(5.00 x 10^3 N/C)$\widehat{x}$ + (6.00 x 10^3 N/C)$\widehat{y}$ + (7.00 x 10^3 N/C)$\widehat{z}$.
Find the electric-charge density distribution p(r) in this space.

Homework Equations

u = (1/2)(ε_o)(E^2) , where ε_o is the constant called the permittivity of free space and = 8.85x10^-12 C^2/N*m^2

I used the equation above, but I was unsure if I was using the correct equation. I thought I maybe should have used one of Maxwell's equations instead: div*E=(ρ)/ε_o

The Attempt at a Solution

magnitude of Electric Field = sqrt((5.00 x 10^3 N/C)² + (6.00 x 10^3 N/C)² + (7.00 x 10^3 N/C)²)

magnitude of E = 10488.1 N/C

u = (1/2)(8.85x10^-12 C^2/N*m^2)(10488.1 N/C)^2
u = 4.87 x 10^-4 C/m^2

Is this the correct way to solve this problem?

 

[Dick]

Homework Statement

A space has an uniform electric field
E=(5.00 x 10^3 N/C)$\widehat{x}$ + (6.00 x 10^3 N/C)$\widehat{y}$ + (7.00 x 10^3 N/C)$\widehat{z}$.
Find the electric-charge density distribution p(r) in this space.

Homework Equations

u = (1/2)(ε_o)(E^2) , where ε_o is the constant called the permittivity of free space and = 8.85x10^-12 C^2/N*m^2

I used the equation above, but I was unsure if I was using the correct equation. I thought I maybe should have used one of Maxwell's equations instead: div*E=(ρ)/ε_o

The Attempt at a Solution

magnitude of Electric Field = sqrt((5.00 x 10^3 N/C)² + (6.00 x 10^3 N/C)² + (7.00 x 10^3 N/C)²)

magnitude of E = 10488.1 N/C

u = (1/2)(8.85x10^-12 C^2/N*m^2)(10488.1 N/C)^2
u = 4.87 x 10^-4 C/m^2

Is this the correct way to solve this problem?

No, u is an energy density. Use the Maxwell equation. What's the divergence of the constant electric field? You know what a divergence is, right?

 

[smantics]

The divergence of a constant electric field is zero, right? I don't understand if the divergence is zero and the divergence is multiplied by the Electric field how you would get a proper answer.

 

[Dick]

The divergence of a constant electric field is zero, right? I don't understand if the divergence is zero and the divergence is multiplied by the Electric field how you would get a proper answer.

Yes, the divergence of a constant E field is zero. The divergence of the field isn't multiplied by the E field. Go back and review 'divergence'. Maxwell's equation is telling you an easy thing about the charge density. What is it?

 

[smantics]

So the divergence of an E field is basically taking partial derivatives with respect to x, y, and z of the E field, but since there are no variables in the E field that means the derivative of the constant E field is zero. Then since the constant E field has a divergence of zero, then does that mean that there cannot be a charge density distribution because the E field doesn't act like a source or a sink, and does that also mean that the charge density is also constant? The concept of divergence is really confusing to me.

 

[Dick]

So the divergence of an E field is basically taking partial derivatives with respect to x, y, and z of the E field, but since there are no variables in the E field that means the derivative of the constant E field is zero. Then since the constant E field has a divergence of zero, then does that mean that there cannot be a charge density distribution because the E field doesn't act like a source or a sink, and does that also mean that the charge density is also constant? The concept of divergence is really confusing to me.

The equation is saying that if the divergence of the E field is zero, then the charge density must be zero, not just a constant, isn't it?

 

[smantics]

So if the E field is zero and the charge density is also zero, and then does it even matter what the E field has numbers? As long as the E field is constant then the charge density will always be zero.

 

[Dick]

So if the E field is zero and the charge density is also zero, and then does it even matter what the E field has numbers? As long as the E field is constant then the charge density will always be zero.

No, the numbers don't matter. Any constant field corresponds to zero charge density.