====== I. Electric charge and field ====== ===== - Electric charge ===== Atoms are the ultimate source electric charge. Atoms have positive protons and neutrals neutrons in their nucleus and negative electrons in a huge cloud surrounding the nucleus. {{:courses:phy102:atom.jpg?nolink|}} ==== - Charging ==== {{:courses:phy102:1-17.jpg?nolink|}} {{:courses:phy102:1-15.jpg?nolink|}} {{:courses:phy102:1-16.jpg?nolink|}} ===== - Coulomb's law ===== {{:courses:phy102:1-2.png?nolink|}} {{:courses:phy102:1-3.jpg?nolink|}} Superposition principle: $\overrightarrow{F}= \sum_n \overrightarrow{F_1}+\overrightarrow{F_2}+...+\overrightarrow{F_n}$ ===== - Electric field ===== {{:courses:phy102:1-4.png?nolink|}} Compare with gravity Vector field (E, g) and scalar field (T). ===== - Electric field from charge ===== From point charges, let us turn to distribution of charges. In this case charge is measured using **charge density**. Charge density can be linear ($\lambda$), surface ($\sigma$) or volume ($\rho$). {{:courses:phy102:1-5.jpg?nolink|}} A differential charge element $dq = \lambda dl$ in one dimension, $\sigma dA$ in 2 dimensions and $\rho dV$ in 3 dimensions. The net electric field at a point $P$ for different charges \begin{align} \vec{E}(P) &= \underbrace{\dfrac{1}{4\pi \epsilon_0} \sum_{i=1}^N \left(\dfrac{q_i}{r^2}\right)\hat{r}}_{\text{Point charges}} \label{eq1}. \end{align} If we have a continuous distribution of charges instead of discrete points, the summation becomes an integration over a line, surface or volume as shown below. \begin{align} \vec{E}(P) &= \underbrace{\dfrac{1}{4\pi \epsilon_0} \int_{line} \left(\dfrac{\lambda \, dl}{r^2}\right) \hat{r}}_{\text{Line charge}} \label{eq2} \\[4pt] \vec{E}(P) &= \underbrace{\dfrac{1}{4\pi \epsilon_0} \int_{surface} \left(\dfrac{\sigma \,dA}{r^2}\right) \hat{r} }_{\text{Surface charge}}\label{eq3} \\[4pt] \vec{E}(P) &= \underbrace{\dfrac{1}{4\pi \epsilon_0} \int_{volume} \left(\dfrac{\rho \,dV}{r^2}\right) \hat{r}}_{\text{Volume charge}} \label{eq4} \end{align} The vector integration indicates that there are in reality three different integrations for the three components $E_x(P)$, $E_y(P)$ and $E_z(P)$. ==== - Line ==== {{:courses:phy102:1-6.png?nolink|}} \begin{align*} \vec{E}(P) &= \vec{E}_1 + \vec{E}_2 \\[4pt] &= E_{1x}\hat{i} + E_{1z}\hat{k} + E_{2x} (-\hat{i}) + E_{2z}\hat{k}. \end{align*} \begin{align*} \vec{E}(P) &= E_{1z}\hat{k} + E_{2z}\hat{k} \\[4pt] &= E_1 \, \cos \, \theta \hat{k} + E_2 \, \cos \, \theta \hat{k}. \end{align*} \begin{align*} \vec{E}(P) &= \dfrac{1}{4 \pi \epsilon_0}\int \dfrac{\lambda dl}{r^2} \, \cos \, \theta \hat{k} + \dfrac{1}{4 \pi \epsilon_0}\int \dfrac{\lambda dl}{r^2} \, \cos \, \theta \hat{k} \\[4pt] &= \dfrac{1}{4 \pi \epsilon_0}\int_0^{L/2} \dfrac{2\lambda dx}{r^2} \, \cos \, \theta \hat{k} \end{align*} Replace $r = (z^2 + x^2)^{1/2}$ and $$\cos \, \theta = \dfrac{z}{r} = \dfrac{z}{(z^2 + x^2)^{1/2}}.$$ \begin{align*} \vec{E}(P) &= \dfrac{1}{4 \pi \epsilon_0}\int_0^{L/2} \dfrac{2\lambda dx}{(z^2 + x^2)} \dfrac{z}{(z^2 + x^2)^{1/2}} \hat{k} \\[4pt] &= \dfrac{1}{4 \pi \epsilon_0}\int_0^{L/2} \dfrac{2\lambda z}{(z^2 + x^2)^{3/2}} dx \hat{k} \\[4pt] &= \dfrac{2 \lambda z}{4 \pi \epsilon_0} \left[\dfrac{x}{z^2\sqrt{z^2 + x^2}}\right]_0^{L/2} \hat{k}. \end{align*} which finally leads to $$ \vec{E}(z) = \dfrac{1}{4 \pi \epsilon_0} \dfrac{\lambda L}{z\sqrt{z^2 + \dfrac{L^2}{4}}} \, \hat{k}. \label{5.12} $$ What happens when $z \gg L$: point charge. If the line extends to infinity, $$ \vec{E}(z) = \dfrac{1}{4 \pi \epsilon_0} \dfrac{2\lambda}{z}\hat{k}. \nonumber$$ ==== - Ring ==== {{:courses:phy102:1-7.png?nolink|}} \begin{align*} \vec{E}(P) &= \dfrac{1}{4\pi \epsilon_0} \int_{line} \dfrac{\lambda dl}{r^2} \hat{r} = \dfrac{1}{4\pi \epsilon_0} \int_0^{2\pi} \dfrac{\lambda Rd\theta}{z^2 + R^2} \dfrac{z}{\sqrt{z^2 + R^2}} \hat{k} \\[4pt] &= \dfrac{1}{4\pi \epsilon_0} \dfrac{\lambda Rz}{(z^2 + R^2)^{3/2}} \hat{k} \int_0^{2\pi} d\theta \\[4pt] &= \dfrac{1}{4\pi \epsilon_0} \dfrac{2\pi \lambda Rz}{(z^2 + R^2)^{3/2}} \hat{k} \\[4pt] &= \dfrac{1}{4\pi \epsilon_0} \dfrac{q_{tot}z}{(z^2 + R^2)^{3/2}} \hat{k}. \end{align*} What happens when $z \gg R$? $$ \vec{E} = \frac{1}{4\pi\epsilon_0} \frac{q_{tot}}{z^2} \hat{k} $$ ==== - Disk ==== {{:courses:phy102:1-8.png?nolink|}} $$ \vec{E}(P) = \dfrac{1}{4\pi \epsilon_0} \int_{surface} \dfrac{\sigma dA}{r^2} \, \cos \, \theta \, \hat{k}. \nonumber $$ Here $dA = 2\pi r'dr'$ and $r^2 = r'^2 + z^2$ and $$ \cos \, \theta = \dfrac{z}{(r'^2 + z^2)^{1/2}}. $$ Substituting all these in the first equation, \begin{align*} \vec{E}(P) &= \vec{E}(z) \\[4pt] &= \dfrac{1}{4 \pi \epsilon_0} \int_0^R \dfrac{\sigma (2\pi r' dr')z}{(r'^2 + z^2)^{3/2}} \hat{k} \\[4pt] &= \dfrac{1}{4 \pi \epsilon_0} (2\pi \sigma z)\left(\dfrac{1}{z} - \dfrac{1}{\sqrt{R^2 + z^2}}\right) \hat{k} \end{align*} If $R\rightarrow \infty$ you get an infinite plane and, \begin{align} \vec{E} &= \lim_{R \rightarrow \infty} \dfrac{1}{4 \pi \epsilon_0} \left( 2 \pi \sigma - \dfrac{2 \pi \sigma z}{\sqrt{R^2 + z^2}}\right)\hat{k} \\[4pt] &= \dfrac{\sigma}{2 \epsilon_0} \hat{k}. \label{5.15} \end{align} ==== - Planes ==== {{:courses:phy102:1-9.png?nolink|}} Electric field points away from positive charges and into the negative charges. Outside, the two planes the two fields cancel each other and inside they add up to give $$ \vec{E} = \dfrac{\sigma}{\epsilon_0}\hat{i}. \nonumber $$ ===== - Field lines ===== {{:courses:phy102:1-10.jpg?nolink|}} {{ :courses:phy102:1-11.jpg?nolink&400 |}} - Electric field lines either originate on positive charges or come in from infinity, and either terminate on negative charges or extend out to infinity. - The number of field lines originating or terminating at a charge is proportional to the magnitude of that charge. A charge of 2q will have twice as many lines as a charge of q. - At every point in space, the field vector at that point is tangent to the field line at that same point. - The field line density at any point in space is proportional to (and therefore is representative of) the magnitude of the field at that point in space. - Field lines can never cross. Since a field line represents the direction of the field at a given point, if two field lines crossed at some point, that would imply that the electric field was pointing in two different directions at a single point. This in turn would suggest that the (net) force on a test charge placed at that point would point in two different directions. Since this is obviously impossible, it follows that field lines must never cross. ===== - Electric dipole ===== {{:courses:phy102:1-12.jpg?nolink|}} If $\vec{d}$ is the distance vector from the negative to the positive charge, the torque \begin{align} \vec{\tau} &= \left(\dfrac{\vec{d}}{2} \times \vec{F}_+ \right) + \left(- \dfrac{\vec{d}}{2} \times \vec{F}_- \right) \\[4pt] &= \left[ \left(\dfrac{\vec{d}}{2}\right) \times \left(+q\vec{E}\right) + \left(-\dfrac{\vec{d}}{2}\right) \times \left(-q\vec{E}\right)\right] \\[4pt] &= q\vec{d} \times \vec{E}. \end{align} where $\vec{p}=q\vec{d}$ is the dipole moment. Hence, $$\vec{\tau} = \vec{p} \times \vec{E}.$$ Let us calculate the electric field of a dipole. {{:courses:phy102:1-15.png?nolink|}} The vertical components of the field at point P cancels out. The horizontal component \begin{align*} E_x &= \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{r^2} \, \sin \, \theta + \dfrac{1}{4\pi \epsilon_0} \dfrac{q}{r^2} \, \sin \, \theta \\[4pt] &= \dfrac{1}{4\pi \epsilon_0} \dfrac{2q}{r^2} \, \sin \, \theta. \end{align*} $$ r^2 = z^2 + \left(\dfrac{d}{2}\right)^2 \nonumber$$ $$ \sin \, \theta = \dfrac{d/2}{R} = \dfrac{d/2}{\left[z^2 + \left(\dfrac{d}{2} \right)^2\right]^{1/2}}. \nonumber $$ So $$ \vec{E}(z) = \dfrac{1}{4\pi \epsilon_0} \dfrac{qd}{\left[z^2 + \left(\dfrac{d}{2}\right)^2\right]^{3/2}} \hat{i}.\label{5.6} $$ If $d \ll z \ll \infty$ then $$ \vec{E}(z) = \dfrac{1}{4\pi \epsilon_0} \dfrac{qd}{z^3}\hat{i}. \nonumber $$ {{:courses:phy102:1-14.jpg?nolink|}} ===== Problems =====