Question-216437

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

Answered by issac last updated on 08/Feb/25

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

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

Answered by MrGaster last updated on 08/Feb/25

(1):▽∙j^→ (∂ρ/∂t)=0⇒▽∙(▽∙B^→ )=μ_0 ▽∙J^→ +μ_0 ε_0 ▽∙(∂E^→ /∂t) ⇒0=μ_0 ▽∙J^→ +μ_0 ε_0 (∂/∂t)(▽∙E^→ ) ▽∙E^→ =(ρ/ε_0 )⇒0=μ_0 ▽∙J^→ +μ_0 (∂ρ/∂t)⇒▽∙j^→ +(∂ρ/∂t)=0 ✓ (2)⇒▽∙B^→ =μ_(0 ) J^→ +μ_0 ε_0 (∂E/∂t) ✓ (3)▽×E=−(∂B^→ /∂t) No! so⇒ determinant ((((2))))

$$\left(\mathrm{1}\right):\bigtriangledown\centerdot\overset{\rightarrow} {{j}}\frac{\partial\rho}{\partial{t}}=\mathrm{0}\Rightarrow\bigtriangledown\centerdot\left(\bigtriangledown\centerdot\overset{\rightarrow} {\boldsymbol{{B}}}\right)=\mu_{\mathrm{0}} \bigtriangledown\centerdot\overset{\rightarrow} {\boldsymbol{{J}}}+\mu_{\mathrm{0}} \epsilon_{\mathrm{0}} \bigtriangledown\centerdot\frac{\partial\overset{\rightarrow} {\boldsymbol{{E}}}}{\partial{t}} \\ $$$$\Rightarrow\mathrm{0}=\mu_{\mathrm{0}} \bigtriangledown\centerdot\overset{\rightarrow} {\boldsymbol{{J}}}+\mu_{\mathrm{0}} \epsilon_{\mathrm{0}} \frac{\partial}{\partial{t}}\left(\bigtriangledown\centerdot\overset{\rightarrow} {\boldsymbol{{E}}}\right) \\ $$$$\bigtriangledown\centerdot\overset{\rightarrow} {{E}}=\frac{\rho}{\epsilon_{\mathrm{0}} }\Rightarrow\mathrm{0}=\mu_{\mathrm{0}} \bigtriangledown\centerdot\overset{\rightarrow} {\boldsymbol{{J}}}+\mu_{\mathrm{0}} \frac{\partial\rho}{\partial{t}}\Rightarrow\bigtriangledown\centerdot\overset{\rightarrow} {{j}}+\frac{\partial\rho}{\partial{t}}=\mathrm{0}\:\checkmark \\ $$$$\left(\mathrm{2}\right)\Rightarrow\bigtriangledown\centerdot\overset{\rightarrow} {\boldsymbol{{B}}}=\mu_{\mathrm{0}\:} \overset{\rightarrow} {\boldsymbol{{J}}}+\mu_{\mathrm{0}} \epsilon_{\mathrm{0}} \frac{\partial\boldsymbol{{E}}}{\partial\boldsymbol{{t}}}\:\checkmark \\ $$$$\left(\mathrm{3}\right)\bigtriangledown×\boldsymbol{{E}}=−\frac{\partial\overset{\rightarrow} {\boldsymbol{{B}}}}{\partial\boldsymbol{{t}}}\:\mathrm{No}! \\ $$$$\mathrm{so}\Rightarrow\begin{array}{|c|}{\left(\mathrm{2}\right)}\\\hline\end{array} \\ $$

Commented by issac last updated on 08/Feb/25