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Question Number 216437 by issac last updated on 08/Feb/25

Answered by issac last updated on 08/Feb/25

$$\mathrm{Q}.$$$$\mathrm{Maxwells}\mathrm{equations}\mathrm{for}\mathrm{electric}\mathrm{and}$$$$\mathrm{magntic}\mathrm{fields}\mathrm{are}\mathrm{as}\mathrm{follows}.$$$$$$$$\mathrm{Choose}\mathrm{one}\mathrm{that}\mathrm{can}\mathrm{be}\mathrm{described}\mathrm{by}\mathrm{the}$$$$\mathrm{above}\mathrm{equation}.$$$$$$$$\mathbf{A})$$$$\mathrm{A}\mathrm{continuous}\mathrm{equation}\mathrm{is}\mathrm{established}$$$$\mathrm{between}\mathrm{charge}\mathrm{density}\mathrm{and}\mathrm{current}$$$$\mathrm{density}$$$$$$$$\mathbf{B})$$$$\mathrm{A}\mathrm{magnetic}\mathrm{field}\mathrm{is}\mathrm{generated}\mathrm{while}\mathrm{the}$$$$\mathrm{capacitor}\mathrm{is}\mathrm{charged}.$$$$$$$$\mathbf{C})$$$$\mathrm{The}\mathrm{strength}\mathrm{of}\mathrm{the}\mathrm{electric}\mathrm{field}$$$$\mathrm{produced}\mathrm{by}\mathrm{a}\mathrm{stationary}\mathrm{point}\mathrm{charge}\mathrm{is}$$$$\mathrm{inversely}\mathrm{proportional}\mathrm{to}\mathrm{the}\mathrm{square}\mathrm{of}\mathrm{the}$$$$\mathrm{distance}$$

Answered by MrGaster last updated on 08/Feb/25

$$\left(1\right):▽\cdot \overrightarrow{j}\frac{\partial \rho }{\partial t}=0\Rightarrow ▽\cdot (▽\cdot \overrightarrow{\mathit{B}})={\mu }_0▽\cdot \overrightarrow{\mathit{J}}+{\mu }_0{ϵ}_0▽\cdot \frac{\partial \overrightarrow{\mathit{E}}}{\partial t}$$$$\Rightarrow 0={\mu }_0▽\cdot \overrightarrow{\mathit{J}}+{\mu }_0{ϵ}_0\frac{\partial }{\partial t}(▽\cdot \overrightarrow{\mathit{E}})$$$$▽\cdot \overrightarrow{E}=\frac{\rho }{{ϵ}_0}\Rightarrow 0={\mu }_0▽\cdot \overrightarrow{\mathit{J}}+{\mu }_0\frac{\partial \rho }{\partial t}\Rightarrow ▽\cdot \overrightarrow{j}+\frac{\partial \rho }{\partial t}=0✓$$$$\left(2\right)\Rightarrow ▽\cdot \overrightarrow{\mathit{B}}={\mu }_0\overrightarrow{\mathit{J}}+{\mu }_0{ϵ}_0\frac{\partial \mathit{E}}{\partial \mathit{t}}✓$$$$\left(3\right)▽\times \mathit{E}=-\frac{\partial \overrightarrow{\mathit{B}}}{\partial \mathit{t}}\mathrm{No}!$$$$\mathrm{so}\Rightarrow \overline{)\begin{array}{c}\left(2\right)\end{array}}$$

Commented by issac last updated on 08/Feb/25

$$\mathrm{yeah}\sim \sim \mathrm{Good}$$

Answered by MathematicalUser2357 last updated on 25/Feb/25

$$\mathrm{Got}$$

Commented by MathematicalUser2357 last updated on 27/Apr/25

Commented by MathematicalUser2357 last updated on 27/Apr/25

"How's the new profile?" BRUH, MATHEMATICIANS ARE NOT RATING THAT. AND, DO NOT BE AN IMAGE

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