Finding magnitude and direction of an electric field

In summary, a spaceship encounters a plane of charged particles with a charge per unit area of σ. The electric field a short distance above the plane is σ/∈0 and is directed perpendicular to the plane. To find the direction, a gaussian surface can be used, such as a short fat cylinder, and it must be considered if all of the flux is directed towards the spacecraft. The orientation of the plane in relation to the spacecraft does not affect the orientation of the electric field.
  • #1
fishturtle1
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Homework Statement


A spaceship encounters a single plane of charged particles, with the charge per unit area equal to σ. The electric field a short distance above the plane has magnitude _____ and is directed _____ to the plane.

a) σ/2∈0, parallel

b) σ/2∈0, perpendicular

c) σ/∈0, parallel

d) σ/∈0, perpendicular

e) 2σ/∈0, parallel

Homework Equations


E = k * (q/r2)

Electric flux = EAcos(φ)

Electric flux = qenclosed / ∈0

The Attempt at a Solution


I'm given σ = q/A

then q = σA

So i plug this into the electric field equation:

E = kσA/r2

substitute k = 1/4π∈0

E = σA/4πr20

I can make a gaussian surface as a sphere, who's surface area is 4πr2 and then substitute that in for A,

then E = σ/∈0

Now I'm not sure how to find the direction and also I'm unsure how to think about this problem. Is the plane under the spaceship or directly in front of it, and what orientation is it at? Am i supposed to find this?
 
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  • #2
fishturtle1 said:
make a gaussian surface as a sphere,
A short fat cylinder (pillbox) would be more usual here, but the result is the same.
You do need to think about whether all of the flux coming out of it is directed towards the spacecraft , though.
fishturtle1 said:
Is the plane under the spaceship or directly in front of it,
It doesn't matter. The question asks for the orientation of the field in relation to that of the plane (not "spaceplane").
 

1. How is the magnitude of an electric field calculated?

The magnitude of an electric field is calculated by dividing the force exerted on a test charge by the magnitude of the test charge itself. This can be represented by the equation E=F/q, where E is the electric field, F is the force, and q is the test charge.

2. What is the unit of measurement for electric field?

The unit of measurement for electric field is Newtons per Coulomb (N/C) in the SI system. In other systems, it may be measured in volts per meter (V/m) or dynes per statcoulomb (dyn/statC).

3. How is the direction of an electric field determined?

The direction of an electric field is determined by the direction of the force it would exert on a positive test charge placed in the field. The field lines always point away from positive charges and towards negative charges.

4. Can the direction of an electric field change?

Yes, the direction of an electric field can change depending on the arrangement of charges in the surrounding area. Electric fields can also be affected by external factors such as conductors or insulators.

5. How is the direction of an electric field represented visually?

The direction of an electric field can be represented visually using field lines. Field lines are drawn to show the direction of the field at different points, with the density of the lines representing the strength of the field. The direction of the field is always tangent to the field lines.

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