Charge over a triangular region

  • Thread starter unknown_2
  • Start date
  • Tags
    Charge
In summary, the formula for calculating the electric field due to a charged triangular region is given by <i>E = (kq)(sinθ1 + sinθ2 + sinθ3)/r^2</i>, where k is the Coulomb's constant, q is the magnitude of the charge, θ1, θ2, and θ3 are the angles between the sides of the triangle and the direction of the electric field, and r is the distance from the charged triangle to the point where the electric field is being calculated. The shape of the triangular region affects the electric field by changing its direction and strength at different points. The magnitude of the charge and the electric field have a direct relationship, and the electric field
  • #1
unknown_2
29
0

Homework Statement


-Given that [tex]\rho_{v}[/tex]=6xy C/[tex]m^{2}[/tex], Calculate the total charge on the triangular region.

-the verticies of the triangle are: (0,0), (2,2) and (4,0)


Homework Equations



[tex]Q=\int_{s}\rho_{v}ds[/tex]

The Attempt at a Solution



[tex]=\int^{4}_{0}\int^{-x+4}_{x}6xy dydx[/tex]
[tex]=\int^{4}_{0}48x - 24x^{2} dx[/tex]
[tex]=24x^{2}-8x^{3}|^{4}_{0}[/tex]
[tex]=-128C[/tex]

According to the textbook the correct answer is 32C

any help would be appreciated!
Thanks,
 
Physics news on Phys.org
  • #2
You got your integration limits mixed up. The maximum value for x is 2 m. Draw a picture and you will see why.
 

1. What is the formula for calculating the electric field due to a charged triangular region?

The formula for calculating the electric field due to a charged triangular region is given by E = (kq)(sinθ1 + sinθ2 + sinθ3)/r^2, where k is the Coulomb's constant, q is the magnitude of the charge, θ1, θ2, and θ3 are the angles between the sides of the triangle and the direction of the electric field, and r is the distance from the charged triangle to the point where the electric field is being calculated.

2. How does the shape of the triangular region affect the electric field it produces?

The shape of the triangular region affects the electric field it produces by changing the direction and strength of the electric field at different points. For example, a sharper angle in the triangle will result in a stronger electric field at that point, while a flatter angle will result in a weaker electric field. The overall shape of the triangle will also determine the direction of the electric field at different points.

3. What is the relationship between the magnitude of the charge and the electric field produced by a triangular region?

The magnitude of the charge and the electric field produced by a triangular region have a direct relationship. This means that as the magnitude of the charge increases, the electric field produced by the triangular region also increases, and vice versa. This relationship is described by the Coulomb's law, which states that the electric field is directly proportional to the magnitude of the charge.

4. Can the electric field produced by a triangular region be negative?

Yes, the electric field produced by a triangular region can be negative. This can happen if the charges in the triangle have opposite signs, resulting in an overall negative charge. In this case, the electric field will point towards the triangle rather than away from it, indicating a negative value.

5. How does the distance from the charged triangular region affect the electric field?

The distance from the charged triangular region has an inverse relationship with the electric field it produces. This means that as the distance increases, the electric field decreases, and vice versa. This is described by the inverse square law, which states that the electric field is inversely proportional to the square of the distance between the charged triangle and the point where the electric field is being calculated.

Similar threads

Replies
11
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
949
  • Topology and Analysis
Replies
4
Views
756
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
2
Views
9K
  • Advanced Physics Homework Help
Replies
9
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
910
  • Advanced Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
2
Replies
64
Views
2K
  • Advanced Physics Homework Help
Replies
15
Views
1K
Back
Top