Electromagnetism Cheatsheet

Cheatsheet Content

### Electric Potential from a Point Charge To find the electric potential $V$ at a point due to a charge $Q$ at the origin, use the formula for the potential of a point charge: $$V = \frac{kQ}{r}$$ #### Steps to calculate: 1. **Identify Given Values:** * **Charge ($Q$):** e.g., $5 \mu C = 5 \times 10^{-6} C$ * **Coulomb's Constant ($k$):** $\approx 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$ * **Coordinates of point:** e.g., $(0, 4, 3) \text{ m}$ 2. **Calculate the Distance ($r$):** The distance from the origin $(0, 0, 0)$ to the point $(0, 4, 3)$ is: $$r = \sqrt{(0-0)^2 + (4-0)^2 + (3-0)^2}$$ $$r = \sqrt{0^2 + 4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ m}$$ 3. **Solve for Potential ($V$):** Substitute the values into the potential formula: $$V = \frac{(9 \times 10^9) \times (5 \times 10^{-6})}{5}$$ $$V = 9 \times 10^3 \text{ V}$$ The electric potential at $(0, 4, 3) \text{ m}$ is $9,000 \text{ V}$ (or $9 \text{ kV}$). ### Force and Electric Field via Superposition To find the force and electric field on a test charge using Coulomb's Law and the principle of superposition. #### 1. Given Data * **Charge 1 ($q_1$):** e.g., $-4 \mu C = -4 \times 10^{-6} \text{ C}$ at $P_1(2, -1, 3) \text{ m}$ * **Charge 2 ($q_2$):** e.g., $4 \mu C = 4 \times 10^{-6} \text{ C}$ at $P_2(2, -1, -3) \text{ m}$ * **Test Charge ($q_0$):** e.g., $1 \mu C = 1 \times 10^{-6} \text{ C}$ at origin $O(0,0,0) \text{ m}$ * **Constant ($k$):** $\approx 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$ #### 2. Electric Field ($\vec{E}$) The electric field at the origin is the vector sum of fields from $q_1$ and $q_2$: $\vec{E} = \vec{E}_1 + \vec{E}_2$. The formula for the electric field at the origin from a charge $q$ at $\vec{r}'$ is $\vec{E} = k \frac{q}{|\vec{r}|^3} \vec{r}$, where $\vec{r}$ is the vector from the charge to the observation point. Here, the observation point is the origin $(0,0,0)$. * **For $q_1$ (at $P_1(2, -1, 3)$):** The vector from $P_1$ to the origin is $\vec{r}_1 = (0-2, 0-(-1), 0-3) = (-2, 1, -3)$. $r_1 = |\vec{r}_1| = \sqrt{(-2)^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$. $$\vec{E}_1 = (9 \times 10^9) \frac{-4 \times 10^{-6}}{(\sqrt{14})^3} (-2\hat{a}_x + \hat{a}_y - 3\hat{a}_z)$$ $$\vec{E}_1 \approx -250.7 (2\hat{a}_x - \hat{a}_y + 3\hat{a}_z) \text{ V/m}$$ *(Correction: The original image had an error in the direction of E1, assuming r1 was (2,-1,3) instead of (-2,1,-3). I am correcting this based on the definition of r as observation point - source charge. The magnitude in the original image calculation seems to be for 4uC, not -4uC, so I will adjust the sign.)* $$\vec{E}_1 = \frac{(9 \times 10^9)(-4 \times 10^{-6})}{14\sqrt{14}} (-2\hat{a}_x + \hat{a}_y - 3\hat{a}_z) \approx -687.1 (-2\hat{a}_x + \hat{a}_y - 3\hat{a}_z)$$ $$\vec{E}_1 \approx 1374.2 \hat{a}_x - 687.1 \hat{a}_y + 2061.3 \hat{a}_z \text{ V/m}$$ * **For $q_2$ (at $P_2(2, -1, -3)$):** The vector from $P_2$ to the origin is $\vec{r}_2 = (0-2, 0-(-1), 0-(-3)) = (-2, 1, 3)$. $r_2 = |\vec{r}_2| = \sqrt{(-2)^2 + 1^2 + 3^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$. $$\vec{E}_2 = (9 \times 10^9) \frac{4 \times 10^{-6}}{(\sqrt{14})^3} (-2\hat{a}_x + \hat{a}_y + 3\hat{a}_z)$$ $$\vec{E}_2 \approx 687.1 (-2\hat{a}_x + \hat{a}_y + 3\hat{a}_z)$$ $$\vec{E}_2 \approx -1374.2 \hat{a}_x + 687.1 \hat{a}_y + 2061.3 \hat{a}_z \text{ V/m}$$ #### Total Electric Field ($\vec{E}$): Summing the components: $$\vec{E} = \vec{E}_1 + \vec{E}_2$$ $$\vec{E} = (1374.2 - 1374.2)\hat{a}_x + (-687.1 + 687.1)\hat{a}_y + (2061.3 + 2061.3)\hat{a}_z$$ $$\vec{E} = 0\hat{a}_x + 0\hat{a}_y + 4122.6 \hat{a}_z \text{ V/m}$$ $$\vec{E} \approx 4122.6 \hat{a}_z \text{ V/m}$$ #### 3. Force ($\vec{F}$) The force on the test charge is calculated using $\vec{F} = q_0 \vec{E}$: $$\vec{F} = (1 \times 10^{-6}) \times (4122.6 \hat{a}_z \text{ V/m})$$ $$\vec{F} \approx 4.1226 \times 10^{-3} \hat{a}_z \text{ N}$$ #### Summary: * **Electric Field:** $\approx 4.12 \hat{a}_z \text{ kV/m}$ * **Force:** $\approx 4.12 \hat{a}_z \text{ mN}$ (milli-Newtons) ### EM Fields in Free Space (Given $\vec{E} = E_m \sin(377t - \beta z) \hat{a}_y$) To find the electromagnetic field quantities in free space given $\vec{E} = E_m \sin(377t - \beta z) \hat{a}_y$, we use Maxwell's equations and constitutive relations. #### 1. Electric Flux Density ($\vec{D}$) In free space, the relationship is $\vec{D} = \epsilon_0 \vec{E}$. $$\vec{D} = \epsilon_0 E_m \sin(377t - \beta z) \hat{a}_y \text{ C/m}^2$$ #### 2. Magnetic Flux Density ($\vec{B}$) Using Faraday's Law, $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$: * Calculate the curl: $$\nabla \times \vec{E} = \begin{vmatrix} \hat{a}_x & \hat{a}_y & \hat{a}_z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & E_m \sin(377t - \beta z) & 0 \end{vmatrix}$$ $$ = \left(\frac{\partial}{\partial y}(0) - \frac{\partial}{\partial z}(E_m \sin(377t - \beta z))\right)\hat{a}_x - \left(\frac{\partial}{\partial x}(0) - \frac{\partial}{\partial z}(0)\right)\hat{a}_y + \left(\frac{\partial}{\partial x}(E_m \sin(377t - \beta z)) - \frac{\partial}{\partial y}(0)\right)\hat{a}_z$$ $$ = -(-\beta E_m \cos(377t - \beta z))\hat{a}_x = \beta E_m \cos(377t - \beta z)\hat{a}_x$$ * Now, $\beta E_m \cos(377t - \beta z)\hat{a}_x = -\frac{\partial \vec{B}}{\partial t}$. * Integrate with respect to time: $$\vec{B} = -\int \beta E_m \cos(377t - \beta z) dt \hat{a}_x$$ $$\vec{B} = -\frac{\beta E_m}{377} \sin(377t - \beta z) \hat{a}_x \text{ T}$$ #### 3. Magnetic Field Intensity ($\vec{H}$) In free space, $\vec{H} = \frac{\vec{B}}{\mu_0}$. $$\vec{H} = -\frac{\beta E_m}{377\mu_0} \sin(377t - \beta z) \hat{a}_x \text{ A/m}$$ ### EM Quantities in Free Space (Given $\vec{E} = E_m \sin(\omega t - \beta z) \hat{a}_y$) To find the requested electromagnetic quantities in free space, given $\vec{E} = E_m \sin(\omega t - \beta z) \hat{a}_y$. #### 1. Electric Flux Density ($\vec{D}$) In free space, $\vec{D} = \epsilon_0 \vec{E}$. $$\vec{D} = \epsilon_0 E_m \sin(\omega t - \beta z) \hat{a}_y$$ #### 2. Displacement Current Density ($\vec{J}_d$) The displacement current density is defined as $\vec{J}_d = \frac{\partial \vec{D}}{\partial t}$: $$\vec{J}_d = \frac{\partial}{\partial t} [\epsilon_0 E_m \sin(\omega t - \beta z)\hat{a}_y]$$ $$\vec{J}_d = \omega \epsilon_0 E_m \cos(\omega t - \beta z)\hat{a}_y$$ #### 3. Magnetic Field Intensity ($\vec{H}$) Using Faraday's Law, $\nabla \times \vec{E} = -\mu_0 \frac{\partial \vec{H}}{\partial t}$: $$\nabla \times \vec{E} = \begin{vmatrix} \hat{a}_x & \hat{a}_y & \hat{a}_z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & E_y & 0 \end{vmatrix} = -\frac{\partial E_y}{\partial z} \hat{a}_x = -(-\beta E_m \cos(\omega t - \beta z)) \hat{a}_x = \beta E_m \cos(\omega t - \beta z) \hat{a}_x$$ Equating terms: $\beta E_m \cos(\omega t - \beta z)\hat{a}_x = -\mu_0 \frac{\partial \vec{H}}{\partial t}$. Integrating with respect to $t$: $$\vec{H} = -\int \frac{\beta E_m}{\mu_0} \cos(\omega t - \beta z) dt \hat{a}_x = -\frac{\beta E_m}{\omega \mu_0} \sin(\omega t - \beta z) \hat{a}_x$$ Using the intrinsic impedance of free space $\eta_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} = \frac{\omega \mu_0}{\beta}$, which implies $\frac{\beta}{\omega \mu_0} = \frac{1}{\eta_0}$. So, this simplifies to: $$\vec{H} = -\frac{E_m}{\eta_0} \sin(\omega t - \beta z) \hat{a}_x$$ #### 4. Magnetic Flux Density ($\vec{B}$) In free space, $\vec{B} = \mu_0 \vec{H}$: $$\vec{B} = -\frac{\mu_0 E_m}{\eta_0} \sin(\omega t - \beta z) \hat{a}_x$$ ### Poisson's and Laplace's Equations To derive Poisson's and Laplace's equations and solve for the potential $V$, we use Gauss' Law in differential form. #### 1. Derivation of Poisson's and Laplace's Equations * **Gauss' Law:** $\nabla \cdot \vec{D} = \rho_v$. In a linear, isotropic, and homogeneous medium, $\vec{D} = \epsilon \vec{E}$. $$\nabla \cdot (\epsilon \vec{E}) = \rho_v \implies \epsilon (\nabla \cdot \vec{E}) = \rho_v \implies \nabla \cdot \vec{E} = \frac{\rho_v}{\epsilon}$$ * **Relation between $\vec{E}$ and $V$:** The electric field is the negative gradient of the potential: $$\vec{E} = -\nabla V$$ * **Poisson's Equation:** Substitute the expression for $\vec{E}$ into Gauss' Law: $$\nabla \cdot (-\nabla V) = \frac{\rho_v}{\epsilon} \implies -\nabla^2 V = \frac{\rho_v}{\epsilon} \implies \nabla^2 V = -\frac{\rho_v}{\epsilon}$$ * **Laplace's Equation:** In a charge-free region where $\rho_v = 0$: $$\nabla^2 V = 0$$ #### 2. Solving for $V$ at $x = 5 \text{ cm}$ **Given Data:** * $V$ is a function of $x$ only: $V(x)$. * Electric field is constant: $\vec{E} = -1.5 \times 10^3 \hat{a}_x \text{ V/m}$. * Boundary condition: At $x = -2 \text{ cm} = -0.02 \text{ m}$, $V = 25 \text{ V}$. **Step-by-Step Solution:** 1. **Relation between $\vec{E}$ and $V$:** Since $\vec{E}$ is in the x-direction only, $E_x = -\frac{dV}{dx}$. $$-1.5 \times 10^3 = -\frac{dV}{dx} \implies \frac{dV}{dx} = 1.5 \times 10^3$$ 2. **Integrate to find $V(x)$:** $$V(x) = \int (1.5 \times 10^3)dx = 1.5 \times 10^3 x + C$$ 3. **Apply Boundary Condition:** At $x = -0.02 \text{ m}$, $V = 25 \text{ V}$: $$25 = (1.5 \times 10^3)(-0.02) + C$$ $$25 = -30 + C \implies C = 55$$ So, the potential equation is: $V(x) = 1500x + 55$. 4. **Calculate $V$ at $x = 5 \text{ cm} = 0.05 \text{ m}$:** $$V(0.05) = 1500(0.05) + 55$$ $$V = 75 + 55 = 130 \text{ V}$$ ### Coulomb's Law in Electrostatics **Statement:** The magnitude of the electrostatic force of attraction or repulsion between two point charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them. **Mathematical Expression:** $$F = k \frac{|Q_1 Q_2|}{r^2}$$ where $k$ is Coulomb's constant ($\approx 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$). #### Force Calculation in Vector Form To calculate the force exerted on $Q_2$ by $Q_1$ ($\vec{F}_{12}$), we use the vector form of Coulomb's Law: $$\vec{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{Q_1 Q_2}{|\vec{R}_{12}|^3} \vec{R}_{12}$$ where $\vec{R}_{12}$ is the vector from $Q_1$ to $Q_2$. #### Steps to calculate: 1. **Identify Given Values:** * $Q_1 = -20 \mu C = -20 \times 10^{-6} \text{ C}$ at $P(-6, 4, 6)$ * $Q_2 = 50 \mu C = 50 \times 10^{-6} \text{ C}$ at $R(5, 8, -2)$ * $\frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$ 2. **Calculate the Position Vector ($\vec{R}_{12}$):** The vector from $Q_1$ to $Q_2$ is: $$\vec{R}_{12} = \vec{R} - \vec{P} = (5 - (-6))\hat{a}_x + (8 - 4)\hat{a}_y + (-2 - 6)\hat{a}_z$$ $$\vec{R}_{12} = 11\hat{a}_x + 4\hat{a}_y - 8\hat{a}_z \text{ m}$$ 3. **Calculate the Magnitude ($|\vec{R}_{12}|$):** $$|\vec{R}_{12}| = \sqrt{11^2 + 4^2 + (-8)^2} = \sqrt{121 + 16 + 64} = \sqrt{201} \approx 14.177 \text{ m}$$ 4. **Substitute into the Vector Formula:** $$\vec{F}_{12} = (9 \times 10^9) \frac{(-20 \times 10^{-6})(50 \times 10^{-6})}{(\sqrt{201})^3} (11\hat{a}_x + 4\hat{a}_y - 8\hat{a}_z)$$ $$\vec{F}_{12} = (9 \times 10^9) \frac{-1000 \times 10^{-12}}{201\sqrt{201}} (11\hat{a}_x + 4\hat{a}_y - 8\hat{a}_z)$$ $$\vec{F}_{12} = \frac{-9 \times 10^{-3}}{201\sqrt{201}} (11\hat{a}_x + 4\hat{a}_y - 8\hat{a}_z)$$ $$\vec{F}_{12} \approx -0.0003158 (11\hat{a}_x + 4\hat{a}_y - 8\hat{a}_z) \text{ N}$$ **Final Result:** $$\vec{F}_{12} \approx -0.00347\hat{a}_x - 0.00126\hat{a}_y + 0.00253\hat{a}_z \text{ N}$$ ### Conduction & Displacement Current Density To find the conduction current density ($\vec{J}_c$) and the displacement current density ($\vec{J}_d$) for a given soil, using physical parameters. #### Given Physical Parameters: * **Conductivity ($\sigma$):** $10^{-3} \text{ S/m}$ * **Relative Permittivity ($\epsilon_r$):** $2.5$ * **Electric Field ($\vec{E}$):** $6.0 \times 10^{-6}\sin(9.0 \times 10^9 t) \hat{a}_x \text{ V/m}$ (Assuming direction along x-axis from context) * **Permittivity of Free Space ($\epsilon_0$):** $\approx 8.854 \times 10^{-12} \text{ F/m}$ #### 1. Conduction Current Density ($\vec{J}_c$) The conduction current density is given by Ohm's law in point form: $$\vec{J}_c = \sigma \vec{E}$$ Substituting the values: $$\vec{J}_c = (10^{-3})[6.0 \times 10^{-6}\sin(9.0 \times 10^9 t)] \hat{a}_x$$ $$\vec{J}_c = 6.0 \times 10^{-9}\sin(9.0 \times 10^9 t) \hat{a}_x \text{ A/m}^2$$ #### 2. Displacement Current Density ($\vec{J}_d$) The displacement current density is defined as the rate of change of the electric flux density ($\vec{D} = \epsilon_0 \epsilon_r \vec{E}$): $$\vec{J}_d = \frac{\partial \vec{D}}{\partial t} = \epsilon_0 \epsilon_r \frac{\partial \vec{E}}{\partial t}$$ First, find the derivative of $\vec{E}$ with respect to $t$: $$\frac{\partial \vec{E}}{\partial t} = \frac{\partial}{\partial t} [6.0 \times 10^{-6}\sin(9.0 \times 10^9 t)] \hat{a}_x$$ $$\frac{\partial \vec{E}}{\partial t} = (6.0 \times 10^{-6})(9.0 \times 10^9)\cos(9.0 \times 10^9 t) \hat{a}_x$$ $$\frac{\partial \vec{E}}{\partial t} = 5.4 \times 10^4 \cos(9.0 \times 10^9 t) \hat{a}_x$$ Now, multiply by the permittivity ($\epsilon = \epsilon_0 \epsilon_r$): $$\vec{J}_d = (8.854 \times 10^{-12})(2.5)(5.4 \times 10^4)\cos(9.0 \times 10^9 t) \hat{a}_x$$ $$\vec{J}_d = (2.2135 \times 10^{-11})(5.4 \times 10^4)\cos(9.0 \times 10^9 t) \hat{a}_x$$ $$\vec{J}_d = 1.19529 \times 10^{-6}\cos(9.0 \times 10^9 t) \hat{a}_x$$ $$\vec{J}_d \approx 1.2 \times 10^{-6}\cos(9.0 \times 10^9 t) \hat{a}_x \text{ A/m}^2$$ ### Electric Field and Potential from Charge at Origin To find the electric field intensity and electric potential at the point $(0, 6, 8) \text{ m}$ due to a charge $Q = -0.2 \mu C$ at the origin. #### 1. Identify the Givens * **Charge ($Q$):** $-0.2 \mu C = -0.2 \times 10^{-6} \text{ C}$ * **Observation point:** $P(0, 6, 8)$ * **Position vector from origin to P:** $\vec{r} = 0\hat{a}_x + 6\hat{a}_y + 8\hat{a}_z$ * **Distance ($r$):** $r = |\vec{r}| = \sqrt{0^2 + 6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ m}$ * **Unit vector ($\hat{a}_r$):** $\hat{a}_r = \frac{\vec{r}}{|\vec{r}|} = \frac{6\hat{a}_y + 8\hat{a}_z}{10} = 0.6\hat{a}_y + 0.8\hat{a}_z$ * **Constant ($k$):** $\frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$ #### 2. Electric Field Intensity ($\vec{E}$) The formula for electric field intensity due to a point charge at the origin is: $$\vec{E} = \frac{kQ}{r^2} \hat{a}_r$$ Substituting the values: $$\vec{E} = \frac{(9 \times 10^9)(-0.2 \times 10^{-6})}{10^2} (0.6\hat{a}_y + 0.8\hat{a}_z)$$ $$\vec{E} = \frac{-1800}{100} (0.6\hat{a}_y + 0.8\hat{a}_z)$$ $$\vec{E} = -18(0.6\hat{a}_y + 0.8\hat{a}_z) \text{ V/m}$$ $$\vec{E} = -10.8\hat{a}_y - 14.4\hat{a}_z \text{ V/m}$$ #### 3. Electric Potential ($V$) The formula for electric potential due to a point charge at the origin is: $$V = \frac{kQ}{r}$$ Substituting the values: $$V = \frac{(9 \times 10^9)(-0.2 \times 10^{-6})}{10}$$ $$V = \frac{-1800}{10}$$ $$V = -180 \text{ V}$$ ### Force Magnitude using Coulomb's Law To find the magnitude of the force on charge $q$, we use Coulomb's Law. $$F = k \frac{|q \cdot Q|}{r^2}$$ #### 1. Identify the Given Values * **Charges:** * $q = -300 \mu C = -300 \times 10^{-6} \text{ C}$ * $Q = 10 \mu C = 10 \times 10^{-6} \text{ C}$ * **Positions:** * $\vec{r}_q = (2, 4, 5) \text{ m}$ * $\vec{r}_Q = (1, 1, 3) \text{ m}$ * **Constants:** * Coulomb's constant $k \approx 8.99 \times 10^9 \text{ N} \cdot \text{m}^2/\text{C}^2$ #### 2. Calculate the Distance ($r$) First, find the displacement vector between the two charges (from $Q$ to $q$): $$\Delta \vec{r} = \vec{r}_q - \vec{r}_Q = (2-1)\hat{a}_x + (4-1)\hat{a}_y + (5-3)\hat{a}_z = 1\hat{a}_x + 3\hat{a}_y + 2\hat{a}_z$$ Now, calculate the magnitude (distance): $$r = |\Delta \vec{r}| = \sqrt{1^2 + 3^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \text{ m}$$ $$r^2 = 14 \text{ m}^2$$ #### 3. Calculate the Force Magnitude ($F$) Plug the values into the formula: $$F = (8.99 \times 10^9) \frac{|(-300 \times 10^{-6}) \cdot (10 \times 10^{-6})|}{14}$$ $$F = (8.99 \times 10^9) \frac{|-3000 \times 10^{-12}|}{14}$$ $$F = (8.99 \times 10^9) \frac{3 \times 10^{-9}}{14}$$ $$F = \frac{26.97}{14}$$ $$F \approx 1.9264 \text{ N}$$ The force magnitude on $q$ is approximately $1.93 \text{ N}$.