Electric field at the center of a hemisphere

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SUMMARY

The electric field at the center of a hemisphere with a uniform surface charge density σ is calculated using the formula E = (σ/2ε0). This conclusion is derived from integrating the contributions of the electric field from the surface charge over the specified bounds. The user initially arrived at an incorrect result of (σ/4ε0) due to a misunderstanding of vector components in the electric field calculation. The correct approach involves recognizing that the electric field is a vector quantity, which has specific directional components that must be accounted for in the integration process.

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  • Understanding of electric fields and surface charge density
  • Familiarity with vector calculus and integration techniques
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  • Proficiency in spherical coordinates and their application in physics
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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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I think you are missing the fact that ##\vec E## is a vector-- it has components (some of which are zero in your exercise)
 

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