> [!summary] The electric field is a vector field of how a positive test charge would experience in every point in space. > **Key equations:** > **Force per charge:** $E=\frac{F}{q}$ > **Electric field for a uniform charged line:** $E = \frac{kQ}{x(x^2+a^2)^{1/2}}$ > **Electric field for a uniform charged ring:** $E = \frac{kQx}{(x^2 + a^2)^{3/2}}$ > **Electric field for a uniform charged disk:** $E = \frac{2kQx}{R^2}(\frac{1}{x}-\frac{1}{\sqrt{x^2 + R^2}})$ >[!info]+ Read Time **⏱ 6 mins** # Definition The electric field is a vector field that tells you the [[Forces|force]] per unit of [[Charge|charge]] a **positive** test [[Charge|charge]] would experience at every point in space. $ \vec{E}=\frac{\vec{F}}{q} $ ![[efield_1.png|400]] [^1] >[!note] Explanation Example of the electric field inside a [[Conductors|conducting sphere.]] # Deriving Electric Fields in a Conductor > [!warning] Assumptions In finding an electric field inside [[Conductors|conductors]], assume the following: > - A [[Batteries|battery]] will create a [[Potential Difference|potential difference]] and create an [[Electric Fields|electric field]], $E = \frac{F}{q}$ > - The [[Electric Potential|electric potential]] is described by $V = \frac{W}{q}$ or $W = Vq$ > - Classical [[Work|work]] is described as $W = F\cdot d$ $ \begin{array}{c} \text{From the definition of work:} \\ (1)\quad W = F\cdot d \\ \\ \text{From electric potential:} \\ (2)\quad W = Vq \\ \\ (1)\Rightarrow (2) \\ Vq = F\cdot d \\ F = \frac{Vq}{d} \\ \\ \text{From our definition of electric field} \\ E = \frac{F}{q} = \frac{V}{d} \end{array} $ The result is an equation to describe the electric field as a change in electric potential over a distance. # Deriving the Electric Field of Uniform Charged Line If we wanted to find the electric field of a uniformly charged line some distance $x$ away. We could find this by using our definition of Electric fields and some integration techniques. >[!warning] Assumptions For this we are going to assume the following: >- Spilt the whole length of $Q$ into two segments $a$ , and $-a$. Where $a$ starts from origin to the positive end of the charged line, and $-a$ starts from origin to the negative end of the charged line >- Because we spilt it into two section the Y components from both ends will cancel out in every point, so we only need to worry about the x direction >- The charge density is $\rho = \frac{dq}{dl}$ and $\rho = Q / 2a$ ![[ef_1.png]] We need to find $dq$ so we find that from $\begin{array}{c} \rho = \frac{dq}{dl} \\ \frac{Q}{2a} = \frac{dq}{dl} \\ \frac{Q}{2a} dl = dq \end{array}$ Now we find a small chunk of $dE$. Note that the $r$ direction is the hypotenuse of the $x$ and $y$ component. $\begin{array}{c} dE = k \frac{dq}{r^2}\\ dE = k \frac{Qdy}{2a(x^2+y^2)} \\ \\ \text{Note because the y direction of dE will always cancel out} \\ \text{We only consider the x compment of dE} \\ \\ dE_x = dEcos(\alpha) \\ dE_x = \frac{kQxdy}{2a(x^2 + y^2)^{3/2}}\\\ E = \int_{-a} ^ a dE_x \\ E = \frac{kQ}{x(x^2+a^2)^{1/2}} \end{array}$ # Deriving the Electric Field Due to a Uniform Charged Ring If we now have a ring of charge $Q$ and want to find a the electric field at some point $x$. We can find this using our definitions of electric fields and some integration. >[!warning] Assumptions To do this assume the following: >- The ring is centred around the axis, so that the electric field at any point in the y-axis will always cancel for charges along the x-axis >- The charges around the ring are uniform >- The charge density ($\rho$) is $\rho = Q / L = Q \ 2\pi a$ if we assume the radius is a >- Note that the charge density is also the rate of chage of our charges over distance. ![[ef_2.png]] Our $dq$ of charges can be solved from knowing the charge density $\begin{array}{c} \rho = \frac{dq}{dl} \\ \rho = \frac{Q}{2\pi a} \\ \\ so: \\ dq = \rho dl \\dq = \frac{Q}{2\pi a} dl \\ \text{Because our charge is a ring} \\ \\ L = a\phi \\ dl = ad\phi \\ dq = \frac{Qad\phi}{2\pi a} \end{array}$ >[!bug] Symmetry Argument Since we now know our $dq$ we can solve for $dE$ in the $x$ direction. Note that the $y$ direction of an electric field will always cancel because we are using the symmetry argument that our ring is centre around the axis. $ \begin{array}{c} dE = k \frac{dq}{r^2} \\ dE = \frac{kQd\phi}{2\pi (x^2 + a^2)} \\ dE_x = dEcos(\theta) \\ dE_x = \frac{kQxd\phi}{2\pi (x^2 + a^2)^{3/2}}\\ \text{Than we intergrate over the ring in term of $d\phi$} \\ E = \int dE_x \\ E = \frac{kQx}{2\pi (x^2 + a^2)^{3/2}}\int_0 ^{2\pi} d\phi \\ E = \frac{kQx}{(x^2 + a^2)^{3/2}} \end{array} $ # Deriving the Electric field for a Uniform Charged Disk If we have a charge uniformly distributed over a disk and want the electric field over some distance along the x-axis, we can again use our definitions from electric field and some integration to solve this. >[!warning] Assumptions Were going to assume the following: >- Our charge density $\rho$ is defined as the Total charge / Total area and is the rate of change of our charge. >- If the disk is centred around the origin, we're going to assume that the electric will cancel out in the y-direction. ![[ef_3.png]] To find $dQ$ we do: $\begin{array}{c} \rho = \frac{dq}{dA} \\ \rho = \frac{Q}{\pi R^2}\\ dq = \rho dA \\ dq = dA \frac{Q}{\pi R^2} \\ \text{Note that $A = \pi r^2 $} \\ dq = \frac{2\pi Q rdr}{\pi R^2} \end{array}$ Then our electric field is, note that r is the hyp distance (from disk to x position), and we will assume it is ($x^2 +y^2$) by pythagorean theorem. $\begin{array}{c} dE = \frac{kdq}{r^2} \\ dE = \frac{kdq}{(x^2+y^2)^2}\\ dE_x = dEcos(\theta) \\ dE_x = \frac{kdq}{(x^2+y^2)^{3/2}} \\ dE_x = \frac{2kxQrdr}{R^2(x^2+y^2)^{3/2}} \\ E = \int dE_x \\ E = \frac{2kQx}{R^2}(\frac{1}{x}-\frac{1}{\sqrt{x^2 + R^2}}) \end{array}$ # Resources <iframe width="560" height="315" src="https://www.youtube.com/embed/bEqe3kaDZKU?si=x2OPMuMz7P21ugPd" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" referrerpolicy="strict-origin-when-cross-origin" allowfullscreen></iframe> --- > 💡 Found this concept helpful? [Star Math & Matter on GitHub](https://github.com/rajeevphysics/Obsidan-MathMatter) to support more intuitive science breakdowns like this. --- [^1]: Taken from https://tikz.net/electric_field_plots/ by Izaak Neutelings (Februari, 2020)