Φ = E → ⋅ A → = E A cos θ = ∫ E → ⋅ d A → = k q r 2 ⋅ 4 π r 2 = q 4 π r 2 4 π ϵ r 2 = q ϵ {\displaystyle \Phi ={\vec {E}}\cdot {\vec {A}}=EA\cos \theta =\int {\vec {E}}\cdot d{\vec {A}}={\frac {kq}{r^{2}}}\cdot 4\pi r^{2}={\frac {\quad q\;\;4\pi r^{2}}{4\pi \epsilon \,r^{2}\quad \;\;}}={\frac {q}{\epsilon }}}
图23.2
E = ∫ d E = k ∫ d q r 2 = k λ ∫ a l + a d x x 2 = k λ [ − 1 x ] a l + a d q = λ d x = k Q l ( − 1 l + a + 1 a ) = k Q a ( l + a ) {\displaystyle {\begin{aligned}E&=\int dE=k\int {\frac {dq}{r^{2}}}=k\lambda \int _{a}^{l+a}{\frac {dx}{x^{2}}}=k\lambda \left[-{\frac {1}{x}}\right]_{a}^{l+a}\\&\qquad \qquad \qquad dq=\lambda dx\\&=k{\frac {Q}{l}}\left(-{\frac {1}{l+a}}+{\frac {1}{a}}\right)={\frac {kQ}{a\left(l+a\right)}}\end{aligned}}}
图23.3
E x = ∫ d E cos θ = ∫ k d q r 2 x r = ∫ k x d q r 3 = k x Q x 2 + a 2 3 {\displaystyle E_{x}=\int dE\,\cos \theta =\int {\frac {kdq}{r^{2}}}{\frac {x}{r}}=\int {\frac {kxdq}{r^{3}}}={\frac {kxQ}{{\sqrt {x^{2}+a^{2}}}^{3}}}}
E x = ∫ k x σ r d r d θ x 2 + r 2 3 = k x 2 π σ ∫ 0 R r d r x 2 + r 2 3 = k x π σ ∫ 0 R d ( x 2 + r 2 ) x 2 + r 2 3 d q = σ d A = σ r d r d θ = 2 π σ r d r = k x π σ ∫ 0 R ( x 2 + r 2 ) − 3 2 d ( x 2 + r 2 ) = − 2 k x π σ [ 1 x 2 + r 2 ] 0 R = − 2 k x π σ ( 1 x 2 + R 2 − 1 x 2 ) = 2 k x π σ ( 1 x − 1 x 2 + R 2 ) = 2 k π σ ( 1 − x x 2 + R 2 ) {\displaystyle {\begin{aligned}E_{x}=\int {\frac {kx\sigma rdrd\theta }{{\sqrt {x^{2}+r^{2}}}^{3}}}=kx2\pi \sigma \int _{0}^{R}{\frac {rdr}{{\sqrt {x^{2}+r^{2}}}^{3}}}=kx\pi \sigma \int _{0}^{R}{\frac {d\left(x^{2}+r^{2}\right)}{{\sqrt {x^{2}+r^{2}}}^{3}}}\\dq=\sigma dA=\sigma rdrd\theta =2\pi \sigma rdr\qquad \qquad \qquad \qquad \qquad \qquad \quad \\=kx\pi \sigma \int _{0}^{R}\left(x^{2}+r^{2}\right)^{-{\frac {3}{2}}}d\left(x^{2}+r^{2}\right)=-2kx\pi \sigma \left[{\frac {1}{\sqrt {x^{2}+r^{2}}}}\right]_{0}^{R}\\=-2kx\pi \sigma \left({\frac {1}{\sqrt {x^{2}+R^{2}}}}-{\frac {1}{\sqrt {x^{2}}}}\right)=2kx\pi \sigma \left({\frac {1}{x}}-{\frac {1}{\sqrt {x^{2}+R^{2}}}}\right)\\=2k\pi \sigma \left(1-{\frac {x}{\sqrt {x^{2}+R^{2}}}}\right)\end{aligned}}}
图23.9
Φ E = 0 {\displaystyle \Phi _{E}=0}
E ( r > a ) = k Q r 2 Φ E = E ∮ d A = Q ϵ = E 4 π r 2 E ( r < a ) = k Q r a 3 E 4 π r 2 = Q i n ϵ = Q r 3 a 3 / ϵ = Q r 3 ϵ a 3 E = k Q i n r 2 Q i n = Q 4 3 π r 3 4 3 π a 3 = Q r 3 a 3 {\displaystyle {\begin{aligned}E\left(r>a\right)={\frac {kQ}{r^{2}}}\\\Phi _{E}=E\oint dA={\frac {Q}{\epsilon }}=E4\pi r^{2}\\E\left(r<a\right)={\frac {kQr}{a^{3}}}\\E4\pi r^{2}={\frac {Q_{in}}{\epsilon }}=Q{\frac {r^{3}}{a^{3}}}{\big /}\epsilon ={\frac {Qr^{3}}{\epsilon a^{3}}}\\E={\frac {kQ_{in}}{r^{2}}}\qquad Q_{in}=Q{\frac {{\frac {4}{3}}\pi r^{3}}{{\frac {4}{3}}\pi a^{3}}}=Q{\frac {r^{3}}{a^{3}}}\end{aligned}}}
E 2 π r l = λ l ϵ E = λ 2 π ϵ r = 2 k λ r {\displaystyle E2\pi rl={\frac {\lambda l}{\epsilon }}\qquad E={\frac {\lambda }{2\pi \epsilon r}}={\frac {2k\lambda }{r}}}
+++++
2 E A = σ A ϵ E = σ 2 ϵ {\displaystyle 2EA={\frac {\sigma A}{\epsilon }}\qquad E={\frac {\sigma }{2\epsilon }}}
![{\displaystyle {\begin{aligned}E&=\int dE=k\int {\frac {dq}{r^{2}}}=k\lambda \int _{a}^{l+a}{\frac {dx}{x^{2}}}=k\lambda \left[-{\frac {1}{x}}\right]_{a}^{l+a}\\&\qquad \qquad \qquad dq=\lambda dx\\&=k{\frac {Q}{l}}\left(-{\frac {1}{l+a}}+{\frac {1}{a}}\right)={\frac {kQ}{a\left(l+a\right)}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc8c745604f5f7ada7c4dba0407bb7f978b3b9ef)
![{\displaystyle {\begin{aligned}E_{x}=\int {\frac {kx\sigma rdrd\theta }{{\sqrt {x^{2}+r^{2}}}^{3}}}=kx2\pi \sigma \int _{0}^{R}{\frac {rdr}{{\sqrt {x^{2}+r^{2}}}^{3}}}=kx\pi \sigma \int _{0}^{R}{\frac {d\left(x^{2}+r^{2}\right)}{{\sqrt {x^{2}+r^{2}}}^{3}}}\\dq=\sigma dA=\sigma rdrd\theta =2\pi \sigma rdr\qquad \qquad \qquad \qquad \qquad \qquad \quad \\=kx\pi \sigma \int _{0}^{R}\left(x^{2}+r^{2}\right)^{-{\frac {3}{2}}}d\left(x^{2}+r^{2}\right)=-2kx\pi \sigma \left[{\frac {1}{\sqrt {x^{2}+r^{2}}}}\right]_{0}^{R}\\=-2kx\pi \sigma \left({\frac {1}{\sqrt {x^{2}+R^{2}}}}-{\frac {1}{\sqrt {x^{2}}}}\right)=2kx\pi \sigma \left({\frac {1}{x}}-{\frac {1}{\sqrt {x^{2}+R^{2}}}}\right)\\=2k\pi \sigma \left(1-{\frac {x}{\sqrt {x^{2}+R^{2}}}}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18ebc4e4bd2b6dd7a8e12c3f9fbce75e6490eb09)
