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Electromagnetism (Vector Analysis (Dot product : \(\underline{a} \cdot…
Electromagnetism
Vector Analysis
Dot product : \(\underline{a} \cdot \underline{b} \equiv|\underline{a}||\underline{b}| \cos \theta \)
Vector (Cross) prodect : \( \underline{a} \times \underline{b} \equiv|\underline{a}||\underline{b}| \sin \theta \hat{\underline{n}} \)
Grad : \( \underline{\nabla} \equiv \hat{x} \frac{\partial}{\partial x}+\hat{y} \frac{\partial}{\partial y}+\hat{z} \frac{\partial}{\partial z} \)
Div : \( \underline{\nabla} \cdot \underline{v}(x, y, z)=\frac{\partial v_{x}(x, y, z)}{\partial x}+\frac{\partial v_{y}(x, y, z)}{\partial y}+\frac{\partial v_{z}(x, y, z)}{\partial z}\)
Gauss's divergence theorem
: \( \iiint_{V}(\operatorname{div} \underline{v}(x, y, z)) d V=\iint_{S_{\text { closed }}} \underline{v}(x, y, z) \cdot d \underline{A} \)
Continuity equation
: \(\underline{\nabla} \cdot(\rho \underline{v})+\frac{\partial \rho}{\partial t}=0\)
curl : \(\operatorname{curl} v(x, y, z) \equiv \nabla \times \underline{v}(x, y, z)\)
Stokes' theorem
: \(\iint_{S_{\text { open }}}(\operatorname{curl} \underline{v}(x, y, z)) \cdot d \underline{A}=\oint_{C} \underline{v}(x, y, z) \cdot d \underline{L} \)
Electrostatics
Coulombs law
: \(\underline{F}=\frac{1}{4 \pi \epsilon_{0}} \frac{q Q}{r^{2}} \hat{\hat{r}}\)
\( \underline{E}(\underline{r})=\frac{F(\underline{r})}{q}=\sum_{i} \frac{1}{4 \pi \epsilon_{0}} \frac{Q_{i}}{r_{i}^{2}} \)
\( \underline{F}=\sum_{i} \underline{F}_{i}=\sum_{i} \frac{1}{4 \pi \epsilon_{0}} \frac{q Q_{i}}{r_{i}^{2}} \stackrel{\hat{r}_{i}}{r_{i}^{2}} \)
\( \underline{F}(\underline{r})=q \underline{E}(\underline{r}) \)
Coulomb's law for charge density
\( \underline{E}(\underline{r})=\iiint_{V} \frac{1}{4 \pi \epsilon_{0}} \frac{\rho\left(R^{\prime}\right)}{r^{\prime 2}} \hat{r}^{\prime} d V^{\prime} \)
Gauss's law
Take flux as \( d\phi_E = \underline{E}(x,y,z).d\underline{A} \)
Flux exiting a closed surface is \( \Phi_E = \frac{1}{\epsilon_0}Q\)
\( \Phi_{E}=\iint_{S_{\text { closed }}} E(x, y, z) . d \underline{A} \)
Combining the two equations gives : \(\iint_{S_{\text { closed }}} \underline{E}(x, y, z) \cdot d \underline{A}=\frac{1}{\epsilon_{0}} Q\)
\( \underline{\nabla} \cdot \underline{E}(x, y, z)=\frac{1}{\epsilon_{0}} \rho(x, y, z) \)
Electric Potential
\( V(\underline{r})=\int_{\underline{r}}^{\infty} \underline{E}\left(\underline{r}^{\prime}\right) \cdot d \underline{r}^{\prime} \)
\( V\left(\underline{r}_{1}\right)-V\left(\underline{r}_{2}\right)=\int_{\underline{r}_{1}}^{\underline{r_{2}}} \underline{E}\left(\underline{r}^{\prime}\right) \cdot d \underline{r}^{\prime}\)
\( \underline{E}(\underline{r}) = - \underline{\nabla}V(\underline{r})\)
\( \oint_{C} \underline{E}(x, y, z) \cdot d \underline{L}=0 \)
\( \underline{\nabla}\times\underline{E}(x,y,z) = 0\)
Magnetostatics
Magnetic forces and fields
\( d^2F \propto \frac{idlIdL}{r^2}\)
\( d^2F = \frac{\mu_0}{4\pi}\frac{idlIdL}{r^2}\)
\( \mu_0\epsilon_0 = \frac{1}{c^2}\)