Element surface
d S = R 2 s i n ( θ ) d θ d α dS=R^2sin(\theta)d\theta d\alpha d S = R 2 s in ( θ ) d θ d α Electric field
E a d d = k σ d S R 2 = k P c o s ( θ ) R 2 s i n ( θ ) d θ d α R 2 = k P c o s ( θ ) s i n ( θ ) d θ d α E_{add}=k\frac{\sigma dS}{R^2}=k\frac{Pcos(\theta)R^2sin(\theta)d\theta d\alpha}{R^2}=kPcos(\theta)sin(\theta)d\theta d\alpha E a dd = k R 2 σ d S = k R 2 P cos ( θ ) R 2 s in ( θ ) d θ d α = k P cos ( θ ) s in ( θ ) d θ d α
E ∣ ∣ = d E a d d c o s ( θ ) = k P c o s 2 ( θ ) s i n ( θ ) d θ d α E_{||}=dE_{add}cos(\theta)=kPcos^2(\theta)sin(\theta)d\theta d\alpha E ∣∣ = d E a dd cos ( θ ) = k P co s 2 ( θ ) s in ( θ ) d θ d α
Integration
E a d d = k P ∫ c o s 2 ( θ ) s i n ( θ ) d θ ∫ d α = 1 3 P ε 0 E_{add}=kP\int cos^2(\theta)sin(\theta)d\theta\int d\alpha=\frac{1}{3}\frac{P}{\varepsilon_0} E a dd = k P ∫ co s 2 ( θ ) s in ( θ ) d θ ∫ d α = 3 1 ε 0 P
Then the electric field in center of the sphere
E ⃗ = E 0 ⃗ + 1 3 P ⃗ ε 0 \vec{E}=\vec{E_0}+\frac{1}{3}\frac{\vec{P}}{\varepsilon_0} E = E 0 + 3 1 ε 0 P
Comments