Electric field at the center of a hemisphere

In summary, the question asks for the electric field at the center of a hemisphere with a radius of a and a uniform surface charge density, bounded by 0<θ<pi/2 and 0<∅<2pi. The equation used to solve for the electric field is (1/4piε0) ∫ p(r') ((r-r')/[r-r']^3) dr'. The attempted solution includes the integration of (r^2)sinθdθd∅ in the direction of r, the constant surface charge density σ, and the magnitude of the electric field, E= (σ/2ε0). However, the solution
  • #1
DoubleE7
1
0

Homework Statement



What is the electric field at the center of a hemisphere bounded by r=a, 0<θ<pi/2 / and 0<∅< 2pi with a uniform surface charge density?

Homework Equations


(1/4piε0) p(r') ((r-r')/[r-r']^3) dr'

The Attempt at a Solution


Forgive my formatting
ds= (r^2)sinθdθd∅ (In direction of r)
p(r')=constant = σ
r=r (in the direction of r)
r'= 0 (I have trouble with understanding r and r')

E= (σ/4piε0) ∫(0 to 2pi) ∫(0 to /2pi) (r^3)sinθdθd∅ / (r^3)
E= (σ/4piε0) (2pi)(1)
E= (σ/2ε0)

Everywhere I look I see the answer to this problem is (σ/4ε0) and I cannot figure out where I am going wrong. I feel as though I am missing to include something in my equation but I am stumped, help would be appreciated.
 
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  • #2
I think you are missing the fact that ##\vec E## is a vector-- it has components (some of which are zero in your exercise)
 

1. What is the formula for calculating the electric field at the center of a hemisphere?

The formula for calculating the electric field at the center of a hemisphere is E = σ/2ε, where σ is the surface charge density and ε is the permittivity of the medium.

2. How do you determine the direction of the electric field at the center of a hemisphere?

The direction of the electric field at the center of a hemisphere is always perpendicular to the surface of the hemisphere, pointing away from it. This can be determined using the right-hand rule, where the thumb points in the direction of the surface charge density and the fingers curl in the direction of the electric field.

3. Can the electric field at the center of a hemisphere be zero?

No, the electric field at the center of a hemisphere cannot be zero. This is because the surface charge density cannot be zero, and the formula for calculating the electric field at the center of a hemisphere does not have any variables that can result in a zero value.

4. How does the electric field at the center of a hemisphere change with the radius of the hemisphere?

The electric field at the center of a hemisphere is inversely proportional to the radius of the hemisphere. This means that as the radius increases, the electric field decreases, and vice versa.

5. Is the electric field at the center of a hemisphere affected by the location of the charge on the hemisphere?

No, the electric field at the center of a hemisphere is not affected by the location of the charge on the hemisphere. As long as the charge is distributed uniformly on the surface of the hemisphere, the electric field at the center will remain the same.

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