>[!summary] Electric potential energy is the potential energy per unit charge. > It is a scalar and doesn't have direction > Electric potential energy refers to how much energy a charge will have to produce in order to get somewhere. Charges always want to have low electric potential energy > **Key equations:** > **Potential energy:** $\Delta V=\frac{\Delta U}{q_0}=-\int_a^b E\cdot ds$ > **Electric field to a single point charge:** $V=\frac{kq}{r}$ > **Electric field inside a conducting sphere:** $E = \frac{Qr}{4\pi R^3 \epsilon_0}$ > **Electric potential inside a conducting sphere:** $V(b) = \frac{Q}{8\pi \epsilon_0 R} (3-\frac{r^2}{R^2})$ # Important Distinction >[!warning] **Not the same as [[Electric Potential]]** Electric potential energy is like gravitational potential energy ([[Gravitational Potential energy]]) It's the potential energy per unit charge (scalar) **doesn't depend of the sign of q** and is to describe the field ([[Charges]]) $\Delta V=\frac{\Delta U}{q_0}=-\int_a^b E\cdot ds$ # What is Electric Potential Energy >[!bug] Distinction The electric potential refers to where a particle naturally wants to be in a electric field and describes the amount of energy to get there **Electric potential is always in reference to some point.** This describes the amount of energy needed to move a charge. If the particle **naturally** want to move (Its electric potential energy is the lowest) **against the path of the field**, we say this is **towards higher potential** If the particle **naturally** wants to be **towards** the **path of the field**, we say it's towards **lower** **potential energy**. ![[ep_2.png]] [^1] >[!note] Explanation Example of a point charge in a electric field. > The path against the electric field is high potential energy since it would **require a lot of energy** to move that charge against the field. > The path towards the electric field is low energy since it would require a **less energy** to move that charge there. ## Potential Due to a Single Point Charge >[!warning] Assumptions The electric potential is to describe the amount of energy the particle has at a given point. > We will assume the following is true: >- From [[Electric Potential]] $U = \frac{kqq_0}{r}$ >- The [[Electric Potential Energy]] is $\Delta V=\frac{\Delta U}{q_0}$ $\begin{array}{c} V = \frac{\Delta U}{q_0} \\ \Delta U = \frac{kq q_0}{r} \\ V=\frac{kq}{r} \end{array}$ # Electric Potential Inside a Conducting Sphere For a conducting sphere we need to assume the following for electric potential. >[!bug] Assumptions >- Our reference point V(a) is at the surface. >- Charges are distributed around the surface but not in the actually conductor. For a conducting sphere all the charge is centred around the surface. The [[Electric Field]] inside the field is always zero because there is no charge enclosed inside the sphere (By [[Gauss's Law]] the E = 0). By the $\Delta V = -\int E \cdot ds$ $V_0$ is always zero in the electric field but because $\Delta V = V_R - V_0$ our $V_R$ will have a value at the surface, and hence is always having the same electric potential. ![[ep_1.png]] [^2] >[!note] Explanation Electric potential for a conducting sphere Note that after were outside the sphere, well assume the sphere as a point mass and so can use $V = kq/r$ # Electric Potential Inside A Non-Conducting Sphere >[!warning] Assumptions For a non-conducintg sphere we will assume the following: >- For a non-conducting sphere, the charges will be distrubted on the charge density throughout. >- To find the [[Electric Field]] we will use [[Gauss's Law]] For a non-conducting sphere the electric field will change depending on how much area you enclose in and how that charge is distrusted. >[!warning] Assumed equations We'll assume the density is $\rho = \frac{Q}{\frac{4}{3}\pi R^3}$ and our volume $V = \frac{4}{3}\pi r^3$ So our Q_enc at any pint is $Q_{enc} = \frac{Qr^3}{R^3}$ So by gausses law our electric field will be: $\begin{array}{c} \oint E \cdot dA = \frac{Q_{enc}}{\epsilon_0} \\ \text{Well assume a spherical area of gassian surfrace} \\ E \cdot 4\pi r^2 = \frac{Q_{enc}}{\epsilon_0} \\ E = \frac{Q_{enc}}{4\pi r^2\epsilon_0} \\ E = \frac{Qr}{4\pi R^3 \epsilon_0} \end{array}$ >[!bug] Assumption To find our electric potential were going to assume that our reference point V(a) at the surface is $V(a) = \frac{Q}{4 \pi \epsilon _0 R}$ so now: $\begin{array}{c} \Delta V = -\int E\cdot ds \\ V(b) -V(a) = -\int E\cdot ds \\ V(b) = V(a) - \int_r ^R \frac{Qr}{4\pi R^3\epsilon _0} dr \\ \text{Intergrate in terms of r } \\ V(b) = \frac{Q}{8\pi \epsilon_0 R} (3-\frac{r^2}{R^2}) \end{array}$ Our electric field will increase as r from the centre and peaks at R, decreasing as $1/r^2$after. [^1]: Taken from https://tikz.net/electric_field/ by Izaak Neutelings (July 2018) [^2]: Taken from https://tikz.net/electric_potential_plots/ by Izaak Neutelings (February 2020)