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V-FdV-V-F-dS-F-2-V-2-dV-V-dS-V-2-dV-V-dS-V-2-dV-V-dS-r-i-q-i-3-r-r-i-2-1-r-r-4pi-3-r-r-

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Question Number 222927 by MrGaster last updated on 11/Jul/25
∫∫∫_V ▽∙FdV=∫∫_∂V F∙dS F=−▽φ ▽∙(−▽φ)=−▽^2 φ ∫∫∫_V (−▽^2 φ)dV=∫∫_∂V (−▽φ)∙dS −∫∫∫_V ▽^2 φdV=−∫∫_∂V ▽φ∙dS ∫∫∫_V ▽^2 φdV=∫∫_∂V ▽φ∙dS ρ(r)=Σ_i q_i δ^((3)) −(r−r_i ) ▽^2 ((1/(∣r−r′∣)))=−4πδ^((3)) (r−r′) φ(r)=∫∫∫_V ((ρ(r′))/(4π∣r−r′∣))dV′ ▽^2 φ=▽^2 ((1/(4π))∫∫∫_V ((ρ(r′))/(∣r−r′∣))dV′) =(1/(4π))∫∫∫_V ρ(r′)▽^2 ((1/(∣r−r′∣)))dV′ =(1/(4π))∫∫∫_V ρ(r′)(−4πδ^((3)) (r−r′))dV′ =−∫∫∫_V ρ(r′)δ^((3)) (r−r′)dV′ =−ρ(r) ▽^2 φ=−ρ
$$\int\int\int_{{V}} \bigtriangledown\centerdot\boldsymbol{\mathrm{F}}{dV}=\int\int_{\partial{V}} \boldsymbol{\mathrm{F}}\centerdot\mathrm{d}\boldsymbol{{S}} \\ $$$$\boldsymbol{\mathrm{F}}=−\bigtriangledown\phi \\ $$$$\bigtriangledown\centerdot\left(−\bigtriangledown\phi\right)=−\bigtriangledown^{\mathrm{2}} \phi \\ $$$$\int\int\int_{{V}} \left(−\bigtriangledown^{\mathrm{2}} \phi\right){dV}=\int\int_{\partial{V}} \left(−\bigtriangledown\phi\right)\centerdot\mathrm{d}\boldsymbol{{S}} \\ $$$$−\int\int\int_{{V}} \bigtriangledown^{\mathrm{2}} \phi\mathrm{d}{V}=−\int\int_{\partial{V}} \bigtriangledown\phi\centerdot{d}\boldsymbol{{S}} \\ $$$$\int\int\int_{{V}} \bigtriangledown^{\mathrm{2}} \phi{dV}=\int\int_{\partial{V}} \bigtriangledown\phi\centerdot{d}\boldsymbol{{S}} \\ $$$$\rho\left(\boldsymbol{{r}}\right)=\underset{{i}} {\sum}{q}_{{i}} \delta^{\left(\mathrm{3}\right)} −\left(\boldsymbol{{r}}−\boldsymbol{{r}}_{{i}} \right) \\ $$$$\bigtriangledown^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mid\boldsymbol{{r}}−\boldsymbol{{r}}'\mid}\right)=−\mathrm{4}\pi\delta^{\left(\mathrm{3}\right)} \left(\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\right) \\ $$$$\phi\left(\boldsymbol{\mathrm{r}}\right)=\int\int\int_{{V}} \frac{\rho\left(\boldsymbol{\mathrm{r}}'\right)}{\mathrm{4}\pi\mid\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\mid}{dV}' \\ $$$$\bigtriangledown^{\mathrm{2}} \phi=\bigtriangledown^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}\pi}\int\int\int_{{V}} \frac{\rho\left(\boldsymbol{\mathrm{r}}'\right)}{\mid\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\mid}{dV}'\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi}\int\int\int_{{V}} \rho\left(\boldsymbol{\mathrm{r}}'\right)\bigtriangledown^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mid\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\mid}\right){dV}' \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}\pi}\int\int\int_{{V}} \rho\left(\boldsymbol{\mathrm{r}}'\right)\left(−\mathrm{4}\pi\delta^{\left(\mathrm{3}\right)} \left(\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\right)\right)\mathrm{d}{V}' \\ $$$$=−\int\int\int_{{V}} \rho\left(\boldsymbol{\mathrm{r}}'\right)\delta^{\left(\mathrm{3}\right)} \left(\boldsymbol{\mathrm{r}}−\boldsymbol{\mathrm{r}}'\right){dV}' \\ $$$$=−\rho\left(\boldsymbol{\mathrm{r}}\right) \\ $$$$\bigtriangledown^{\mathrm{2}} \phi=−\rho \\ $$

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