U r = k q Q r V r = U r q = k Q r V p = k ∫ d q r Δ V = − ∫ 1 2 E → d S → E → = − ∂ V ∂ x i ^ {\displaystyle {\begin{aligned}U_{r}={\frac {kqQ}{r}}\qquad V_{r}={\frac {U_{r}}{q}}={\frac {kQ}{r}}\qquad V_{p}=k\int {\frac {dq}{r}}\qquad \Delta V=-\int _{1}^{2}{\vec {E}}d{\vec {S}}\\{\vec {E}}={\frac {-\partial V}{\partial x}}{\hat {i}}\end{aligned}}}
絕緣體球導體球
V = k ∫ d q r = k Q x 2 + a 2 E x = − d V d x = k x Q x 2 + a 2 3 {\displaystyle {\begin{aligned}V=k\int {\frac {dq}{r}}={\frac {kQ}{\sqrt {x^{2}+a^{2}}}}\\E_{x}={\frac {-dV}{dx}}={\frac {kxQ}{{\sqrt {x^{2}+a^{2}}}^{3}}}\end{aligned}}}
V = ∫ d V = ∫ k d q x 2 + r 2 = ∫ k 2 π σ r d r x 2 + r 2 = k 2 π σ ∫ 0 R r d r x 2 + r 2 = k 2 π σ ∫ 0 R 1 2 r r ( x 2 + r 2 ) − 1 2 d ( x 2 + r 2 ) = k π σ 2 [ ( x 2 + r 2 ) 1 2 ] 0 R = 2 k π σ ( x 2 + R 2 − x ) E x = − d V d x = − 2 k π σ d [ ( x 2 + R 2 ) 1 2 − x ] d x = − 2 k π σ ( 2 x 2 1 x 2 + R 2 − 1 ) = 2 k π σ ( 1 − x x 2 + R 2 ) {\displaystyle {\begin{aligned}V=\int dV=\int {\frac {kdq}{\sqrt {x^{2}+r^{2}}}}=\int {\frac {k2\pi \sigma rdr}{\sqrt {x^{2}+r^{2}}}}=k2\pi \sigma \int _{0}^{R}{\frac {rdr}{\sqrt {x^{2}+r^{2}}}}\\=k2\pi \sigma \int _{0}^{R}{\frac {1}{2r}}r\left(x^{2}+r^{2}\right)^{\frac {-1}{2}}d\left(x^{2}+r^{2}\right)=k\pi \sigma 2\left[(x^{2}+r^{2})^{\frac {1}{2}}\right]_{0}^{R}\\=2k\pi \sigma ({\sqrt {x^{2}+R^{2}}}-x)\\E_{x}={\frac {-dV}{dx}}=-2k\pi \sigma {\frac {d\left[(x^{2}+R^{2})^{\frac {1}{2}}-x\right]}{dx}}=-2k\pi \sigma \left({\frac {2x}{2}}{\frac {1}{\sqrt {x^{2}+R^{2}}}}-1\right)\\=2k\pi \sigma \left(1-{\frac {x}{\sqrt {x^{2}+R^{2}}}}\right)\end{aligned}}}
V p = k ∫ d q r = k ∫ 0 l λ d x x 2 + a 2 = k λ ∫ 0 l d x x 2 + a 2 ∵ ∫ d x x 2 + a 2 = ln ( x + x 2 + a 2 a ) + C ∴ V p = k λ [ ln ( x + x 2 + a 2 a ) ] 0 l = k λ [ ln ( l + l 2 + a 2 a ) − ln a 2 a ] = k λ ln ( l + l 2 + a 2 a ) {\displaystyle {\begin{aligned}V_{p}=k\int {\frac {dq}{r}}=k\int _{0}^{l}{\frac {\lambda dx}{\sqrt {x^{2}+a^{2}}}}=k\lambda \int _{0}^{l}{\frac {dx}{\sqrt {x^{2}+a^{2}}}}\\\because \int {\frac {dx}{\sqrt {x^{2}+a^{2}}}}=\ln \left({\frac {x+{\sqrt {x^{2}+a^{2}}}}{a}}\right)+{\text{C}}\\\therefore V_{p}=k\lambda \left[\ln \left({\frac {x+{\sqrt {x^{2}+a^{2}}}}{a}}\right)\right]_{0}^{l}\\=k\lambda \left[\ln \left({\frac {l+{\sqrt {l^{2}+a^{2}}}}{a}}\right)-\ln {\frac {\sqrt {a^{2}}}{a}}\right]\\=k\lambda \ln \left({\frac {l+{\sqrt {l^{2}+a^{2}}}}{a}}\right)\end{aligned}}}
導體的靜電平衡有以下性質：
λ = a sin θ d E = k d q r 2 d q = λ d s = a sin θ r d θ d E x = d E cos θ = k d q cos θ r 2 = k a sin θ cos θ d θ r d E y = d E sin θ = k d q sin θ r 2 = k a sin 2 θ d θ r E x = k a r ∫ θ 1 θ 2 sin θ cos θ d θ = k a 4 r ∫ θ 1 θ 2 sin 2 θ d 2 θ = − k a 4 r ( cos 2 θ 2 − cos 2 θ 1 ) E y = k a r ∫ θ 1 θ 2 sin 2 θ d θ = k a 4 r ∫ θ 1 θ 2 ( 1 − cos 2 θ ) d 2 θ = k a 4 r [ 2 θ − sin 2 θ ] θ 1 θ 2 = k a 4 r ( 2 θ 2 − 2 θ 1 − sin 2 θ 2 + sin 2 θ 1 ) E → = E x i ^ − E y j ^ = − k a 4 r ( cos 2 θ 2 − cos 2 θ 1 ) i ^ − k a 4 r ( 2 θ 2 − 2 θ 1 − sin 2 θ 2 + sin 2 θ 1 ) j ^ V x = − ∫ E x d S = − ∫ E x r d θ = k a 8 ∫ θ 2 θ 1 cos 2 θ d 2 θ = k a 8 [ sin 2 θ ] θ 2 θ 1 = k a 8 ( sin 2 θ 2 − sin 2 θ 1 ) V y = − ∫ E y d S = − ∫ E y r d θ = − k a 4 r ∫ θ 1 θ 2 2 θ − sin 2 θ d θ = − k a 4 r [ θ 2 + cos 2 θ 2 ] θ 1 θ 2 {\displaystyle {\begin{aligned}&\lambda =a\sin \theta \\&dE={\frac {kdq}{r^{2}}}\\&dq=\lambda ds=a\sin \theta rd\theta \\&dE_{x}=dE\cos \theta ={\frac {kdq\cos \theta }{r^{2}}}={\frac {ka\sin \theta \cos \theta d\theta }{r}}\\&dE_{y}=dE\sin \theta ={\frac {kdq\sin \theta }{r^{2}}}={\frac {ka\sin ^{2}\theta d\theta }{r}}\\&E_{x}={\frac {ka}{r}}\int _{\theta _{1}}^{\theta _{2}}\sin \theta \cos \theta d\theta ={\frac {ka}{4r}}\int _{\theta _{1}}^{\theta _{2}}\sin 2\theta d2\theta \\&={\frac {-ka}{4r}}(\cos 2\theta _{2}-\cos 2\theta _{1})\\&E_{y}={\frac {ka}{r}}\int _{\theta _{1}}^{\theta _{2}}\sin ^{2}\theta d\theta ={\frac {ka}{4r}}\int _{\theta _{1}}^{\theta _{2}}\left(1-\cos 2\theta \right)d2\theta \\&={\frac {ka}{4r}}{\bigg [}2\theta -\sin 2\theta {\bigg ]}_{\theta _{1}}^{\theta _{2}}={\frac {ka}{4r}}\left(2\theta _{2}-2\theta _{1}-\sin 2\theta _{2}+\sin 2\theta _{1}\right)\\&{\vec {E}}=E_{x}{\hat {i}}-E_{y}{\hat {j}}\\&={\frac {-ka}{4r}}\left(\cos 2\theta _{2}-\cos 2\theta _{1}\right){\hat {i}}-{\frac {ka}{4r}}\left(2\theta _{2}-2\theta _{1}-\sin 2\theta _{2}+\sin 2\theta _{1}\right){\hat {j}}\\&V_{x}=-\int E_{x}dS=-\int E_{x}rd\theta \\&={\frac {ka}{8}}\int _{\theta _{2}}^{\theta _{1}}\cos 2\theta d2\theta ={\frac {ka}{8}}{\bigg [}\sin 2\theta {\bigg ]}_{\theta _{2}}^{\theta _{1}}\\&={\frac {ka}{8}}\left(\sin 2\theta _{2}-\sin 2\theta _{1}\right)\\&V_{y}=-\int E_{y}dS=-\int E_{y}rd\theta \\&={\frac {-ka}{4r}}\int _{\theta _{1}}{\theta _{2}}2\theta -\sin 2\theta d\theta \\&={\frac {-ka}{4r}}{\bigg [}\theta ^{2}+{\frac {\cos 2\theta }{2}}{\bigg ]}_{\theta _{1}}^{\theta _{2}}\end{aligned}}}
![{\displaystyle {\begin{aligned}V=\int dV=\int {\frac {kdq}{\sqrt {x^{2}+r^{2}}}}=\int {\frac {k2\pi \sigma rdr}{\sqrt {x^{2}+r^{2}}}}=k2\pi \sigma \int _{0}^{R}{\frac {rdr}{\sqrt {x^{2}+r^{2}}}}\\=k2\pi \sigma \int _{0}^{R}{\frac {1}{2r}}r\left(x^{2}+r^{2}\right)^{\frac {-1}{2}}d\left(x^{2}+r^{2}\right)=k\pi \sigma 2\left[(x^{2}+r^{2})^{\frac {1}{2}}\right]_{0}^{R}\\=2k\pi \sigma ({\sqrt {x^{2}+R^{2}}}-x)\\E_{x}={\frac {-dV}{dx}}=-2k\pi \sigma {\frac {d\left[(x^{2}+R^{2})^{\frac {1}{2}}-x\right]}{dx}}=-2k\pi \sigma \left({\frac {2x}{2}}{\frac {1}{\sqrt {x^{2}+R^{2}}}}-1\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/0cb282ae783fe29bd5e210c8adbd3853f40465a4)
![{\displaystyle {\begin{aligned}V_{p}=k\int {\frac {dq}{r}}=k\int _{0}^{l}{\frac {\lambda dx}{\sqrt {x^{2}+a^{2}}}}=k\lambda \int _{0}^{l}{\frac {dx}{\sqrt {x^{2}+a^{2}}}}\\\because \int {\frac {dx}{\sqrt {x^{2}+a^{2}}}}=\ln \left({\frac {x+{\sqrt {x^{2}+a^{2}}}}{a}}\right)+{\text{C}}\\\therefore V_{p}=k\lambda \left[\ln \left({\frac {x+{\sqrt {x^{2}+a^{2}}}}{a}}\right)\right]_{0}^{l}\\=k\lambda \left[\ln \left({\frac {l+{\sqrt {l^{2}+a^{2}}}}{a}}\right)-\ln {\frac {\sqrt {a^{2}}}{a}}\right]\\=k\lambda \ln \left({\frac {l+{\sqrt {l^{2}+a^{2}}}}{a}}\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2224500607f9b2cb1c6ba0c14142cb0d4c1ea873)
![{\displaystyle {\begin{aligned}&\lambda =a\sin \theta \\&dE={\frac {kdq}{r^{2}}}\\&dq=\lambda ds=a\sin \theta rd\theta \\&dE_{x}=dE\cos \theta ={\frac {kdq\cos \theta }{r^{2}}}={\frac {ka\sin \theta \cos \theta d\theta }{r}}\\&dE_{y}=dE\sin \theta ={\frac {kdq\sin \theta }{r^{2}}}={\frac {ka\sin ^{2}\theta d\theta }{r}}\\&E_{x}={\frac {ka}{r}}\int _{\theta _{1}}^{\theta _{2}}\sin \theta \cos \theta d\theta ={\frac {ka}{4r}}\int _{\theta _{1}}^{\theta _{2}}\sin 2\theta d2\theta \\&={\frac {-ka}{4r}}(\cos 2\theta _{2}-\cos 2\theta _{1})\\&E_{y}={\frac {ka}{r}}\int _{\theta _{1}}^{\theta _{2}}\sin ^{2}\theta d\theta ={\frac {ka}{4r}}\int _{\theta _{1}}^{\theta _{2}}\left(1-\cos 2\theta \right)d2\theta \\&={\frac {ka}{4r}}{\bigg [}2\theta -\sin 2\theta {\bigg ]}_{\theta _{1}}^{\theta _{2}}={\frac {ka}{4r}}\left(2\theta _{2}-2\theta _{1}-\sin 2\theta _{2}+\sin 2\theta _{1}\right)\\&{\vec {E}}=E_{x}{\hat {i}}-E_{y}{\hat {j}}\\&={\frac {-ka}{4r}}\left(\cos 2\theta _{2}-\cos 2\theta _{1}\right){\hat {i}}-{\frac {ka}{4r}}\left(2\theta _{2}-2\theta _{1}-\sin 2\theta _{2}+\sin 2\theta _{1}\right){\hat {j}}\\&V_{x}=-\int E_{x}dS=-\int E_{x}rd\theta \\&={\frac {ka}{8}}\int _{\theta _{2}}^{\theta _{1}}\cos 2\theta d2\theta ={\frac {ka}{8}}{\bigg [}\sin 2\theta {\bigg ]}_{\theta _{2}}^{\theta _{1}}\\&={\frac {ka}{8}}\left(\sin 2\theta _{2}-\sin 2\theta _{1}\right)\\&V_{y}=-\int E_{y}dS=-\int E_{y}rd\theta \\&={\frac {-ka}{4r}}\int _{\theta _{1}}{\theta _{2}}2\theta -\sin 2\theta d\theta \\&={\frac {-ka}{4r}}{\bigg [}\theta ^{2}+{\frac {\cos 2\theta }{2}}{\bigg ]}_{\theta _{1}}^{\theta _{2}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f87e97c0f975455fc268106e41c57dbb5ebd3702)
