Electric Force & Electric Field
Electric Field Definition
By experiment, we know that the force acted on a test charge \(Q\) at \(\bf{r_{1}}\) by a single point charge \(q\) at \(\bf{r_2}\) is
$$\textbf{F} = { \frac{1}{4\pi\epsilon_0} \frac{qQ}{r^2} \hat{\textbf{r}}}.$$
where \(\textbf{r} = \textbf{r}_2 - \textbf{r}_1 \). This is known as the Coulomb's law.
For more than one charge, the electric force could be added
$$\textbf{F} = { \frac{1}{4\pi\epsilon_0} \sum_{i=1}^{n} \frac{q_i Q}{r_i^2} \hat{\textbf{r}_i}}.$$
One can define the electric field as
$$\textbf{E} = { \frac{1}{4\pi\epsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_i^2} \hat{\textbf{r}_i}}.$$
or for continuous charge distribution
$$\textbf{E} = { \frac{1}{4\pi\epsilon_0} \int \frac{1}{r^2} \hat{\textbf{r}} dq }.$$
which is the force would be added by the charge distribution per unit charge.
Find the electric field a distance \(z\) above the midpoint of a straight line segment of length \(2L\), which carries a uniform line charge \(\lambda\)
Solution: $$ d\vec{E} = 2\frac{1}{4\pi\epsilon_0}\left(\frac{\lambda dx}{d^2}\right)\cos\theta \hat{z}$$ where \(\cos\theta=z/d\) and \(d=\sqrt{z^2+x^2}\) and \(0\leq x \leq L\) \begin{align} E &= \frac{1}{4\pi\epsilon_0}\int^L_0\frac{2\lambda z}{(z^2+x^2)^{3/2}}dx\\ &= \frac{2\lambda z}{4\pi\epsilon_0}\left[\frac{x}{z^2\sqrt{z^2+x^2}}\right]^L_0\\ &= \frac{1}{4\pi\epsilon_0}\frac{2\lambda L}{z\sqrt{z^2+L^2}} \end{align}
Find the electric field a distance \(z\) above one end of a straight line segment of length \(L\), which carries a uniform line charge \(\lambda\).
$$E_z = \frac{1}{4\pi\epsilon_0}\int^L_0 \frac{\lambda dx}{d^2}\cos\theta$$ where \(d^2=z^2+x^2\) and \(\cos\theta=\frac{z}{d}\) \begin{align} E_z &= \frac{1}{4\pi\epsilon_0}\lambda z\int^L_0\frac{1}{(z^2+x^2)^{3/2}}dx \\ &= \frac{1}{4\pi\epsilon_0}\lambda z \left[\frac{1}{z^2}\frac{x}{\sqrt{z^2+x^2}}\right]^L_0\\ &= \frac{1}{4\pi\epsilon_0}\frac{\lambda}{z}\frac{L}{\sqrt{z^2+L^2}} \end{align} \begin{align} E_x &= -\frac{1}{4\pi\epsilon_0}\int^L_0 \frac{\lambda dx}{d^2}\sin\theta \\ &= -\frac{1}{4\pi\epsilon_0}\lambda \int \frac{xdx}{(x^2+z^2)^{3/2}}\\ &= -\frac{1}{4\pi\epsilon_0}\left[-\frac{1}{\sqrt{x^2+z^2}}\right]^L_0\\ &= -\frac{1}{4\pi\epsilon_0} \lambda\left[\frac{1}{z} - \frac{1}{\sqrt{z^2+L^2}}\right] \end{align}
Divergence of Electric Field
The flux of an electric field through a surface \(S\) is defined as $$\Phi = { \int _{S}\textbf{E} \cdot d\textbf{a} }.$$ to describe the "amount" of electric field passing through the surface \(S\). \begin{align} \textbf{E} & = { \frac{1}{4\pi\epsilon_0} \int \frac{1}{r^2} \hat{\textbf{r}} dq } \\ \nabla \cdot \textbf{E} &= { \frac{1}{4\pi\epsilon_0} \int \nabla \cdot ( \frac{1}{r^2} \hat{\textbf{r}} ) dq }\\ &= { \frac{1}{4\pi\epsilon_0} \int (4 \pi) \delta (\textbf{r}) dq }\\ &= { \frac{1}{4\pi\epsilon_0} \int \rho (4 \pi) \delta (\textbf{r}) d\tau }\\ &= { \frac{1}{\epsilon_0} \rho(\textbf{r}) }\\ \end{align} Integrate around a volume, \begin{align} \int \nabla \cdot \textbf{E} d\tau &= \int { \frac{1}{\epsilon_0} \rho(\textbf{r}) } d\tau \\ \end{align} By divergence theorem, \begin{align} \int \nabla \cdot \textbf{E} d\tau = \int_{S} \textbf{E} \cdot d\textbf{a} &= \int { \frac{1}{\epsilon_0} \rho(\textbf{r}) } d\tau \\ \int_{S} \textbf{E} \cdot d\textbf{a} &= \frac{Q_{enc}}{\epsilon_0} \\ \end{align} where \(Q_{enc}\) is the charge enclosed by the surface \(S\). So, we have $$ \bbox[5px,border:2px solid #666] { \nabla \cdot \textbf{E} = { \frac{1}{\epsilon_0} \rho(\textbf{r}) } } $$ or equivalently $$ \bbox[5px,border:2px solid #666] { \int_{S} \textbf{E} \cdot d\textbf{a} = \frac{Q_{enc}}{\epsilon_0} } $$ This is called the Gauss's Law
Find the electric field outside a uniformly charged solid sphere of radius \(R\) and total charge \(q\)
Solution: \begin{align} \oint_S \vec{E}\cdot d\vec{a} &= \frac{1}{\epsilon_0}Q_{\text{enc}}\\ |\vec{E}|4\pi r^2 &= \frac{1}{\epsilon_0}q\\ \vec{E} &= \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\hat{r} \end{align}
A hollow spherical shell carries charge density $$\rho = \frac{k}{r^2}$$ int the region \(a\leq r \leq b\). Find the electric field in the three regions: (i) \( r<a \), (ii) \( a<r<b \), (iii)\( r>b \).
\( r<a \)
\(Q_{\text{enc}}=0\), so \(\vec{E}=0\)
\( a<r<b \)
\begin{align}
\oint \vec{E}\cdot d\vec{a} &= \frac{1}{\epsilon_0}Q_{\text{enc}}\\
E(4\pi r^2) &= \frac{1}{\epsilon_0}\int\rho d\tau \\
&= \frac{1}{\epsilon_0}\int^{2\pi}_0\int^{\pi}_0\int^r_a\frac{k}{r^2}r^2\sin\theta dr d\theta d\phi \\
&= \frac{4\pi k}{\epsilon_0}\int^r_a dr \\
&= \frac{4 \pi k}{\epsilon_0}(r-a)\\
\end{align}
So,
$$\vec{E} = \frac{k}{\epsilon_0}\left(\frac{r-a}{r^2}\right)\hat{r}$$
\( r>b \)
$$E(4\pi r^2) = \frac{4\pi k}{\epsilon_0}\int^b_a dr = \frac{4\pi k}{\epsilon_0}(b-a)$$
so,
$$\vec{E} = \frac{k}{\epsilon_0}\left(\frac{b-a}{r^2}\right)\hat{r}$$
Curl of Electric Field
\begin{align} \textbf{E} &= { \frac{1}{4\pi\epsilon_0} \int \frac{1}{r^2} \hat{\textbf{r}} dq } \\ \nabla \times \textbf{E} &= { \frac{1}{4\pi\epsilon_0} \int \nabla \times (\frac{\rho}{r^2} \hat{\textbf{r}} ) d\tau } \\ &= { \frac{1}{4\pi\epsilon_0} \int (\nabla \times (\frac{\hat{\textbf{r}}}{r^2})) \rho d\tau } \\ &= 0 \end{align} So, electric field is curl-free. $$ \bbox[5px,border:2px solid #666] { \nabla \times \textbf{E} = 0 } $$ Because of this, the line integral of electric field is path independent, i.e. between 2 points, no matter which path is used for the line integral, the line integral result would be the same. Hence, one could define something called electric potential.
Electric Potential
The electric potential from a reference point \(O\) to a point \(\textbf{r}\) is defined as
$$ V(\textbf{r})= {- \int_{O}^{\textbf{r}} \textbf{E}(\textbf{r}) \cdot d\textbf{l}} $$
It is the work done to a unit charge from the reference point \(O\) to the point \(\textbf{r}\) in a way such that no kinetic energy is added to the unit charge, so the force would be exactly \(-\textbf{E}\).
The potential difference between two points, \(a\) and \(b\), is the difference of the electric potential at \(a\) and \(b\).
\begin{align}
V(\textbf{b}) - V(\textbf{a}) &= {- \int_{O}^{\textbf{b}} \textbf{E}(\textbf{r}) \cdot d\textbf{l} - (- \int_{O}^{\textbf{a}} \textbf{E}(\textbf{r}) \cdot d\textbf{l}) } \\
\ &= - \int_{\textbf{a}}^{\textbf{b}} \textbf{E}(\textbf{r}) \cdot d\textbf{l} \\
\end{align}
By fundamental theorem for gradients,
$$ V(\textbf{b}) - V(\textbf{a}) = - \int_{\textbf{a}}^{\textbf{b}} \nabla V(\textbf{r}) \cdot d\textbf{l} $$
So,
$$
\textbf{E} = -\nabla V
$$
By the definition of electric potential, setting the infinity as the reference point, the potential at \(\textbf{r}_1\) by a single point charge \(q\)
\begin{align}
V(\textbf{r}_1) &= {- \int_{-\infty}^{\textbf{r}_1} \textbf{E}(\textbf{r}_1) \cdot d\textbf{l}} \\
&= {- \int_{-\infty}^{\textbf{r}_1} \frac{1}{4\pi\epsilon_0} \frac{q}{r^2} \hat{\textbf{r}} \cdot d\textbf{r}} \\
&= {\frac{1}{4\pi\epsilon_0} \frac{q}{r}} \\
\end{align}
For multiple point charges, the potential became
$$V = { \frac{1}{4\pi\epsilon_0} \sum_{i=1}^{n} \frac{q_i}{r} }.$$
or for continuous charge distribution
$$V = { \frac{1}{4\pi\epsilon_0} \int \frac{1}{r} dq }.$$
which is only true if the charge distribution \(dq\) is independent of \(r\) as it has integrated \(\textbf{E}\cdot d\textbf{r}\) to give \(\frac{1}{r}\)
Find the potential of a uniformly charged spherical shell of radius \(R\)
Solution: $$V(\vec{r}) = \frac{1}{4\pi\epsilon_0}\int\frac{\sigma}{d}da$$ where \(d^2 = R^2 + z^2 -2Rz\cos\theta\) \begin{align} 4\pi\epsilon_0 V(z) &= \sigma \int^{2\pi}_0\int^{\pi}_0\frac{R^2\sin\theta}{\sqrt{R^2+z^2-2Rz\cos\theta}}d\theta d\phi \\ &= 2\pi R^2\sigma \int^{\pi}_0 \frac{\sin\theta}{\sqrt{R^2+z^2-2Rz\cos\theta}}d\theta \\ &= 2\pi R^2\sigma\left[\frac{1}{Rz}\sqrt{R^2+z^2-2Rz\cos\theta}\right]^{\pi}_0\\ &= \frac{2\pi R\sigma}{z}\left(\sqrt{R^2 + z^2 + 2Rz}-\sqrt{R^2+z^2-2Rz}\right)\\ &= \frac{2\pi R\sigma}{z}\left(\sqrt{(R+z)^2}-\sqrt{(R-z)^2}\right) \end{align} When \(z\) is outside the sphere, \(z-R > 0\), so \(\sqrt{(R-z)^2}=z-R\) $$V(z) = \frac{R\sigma}{2\epsilon_0 z}\left[(R+z)-(z-R)\right] = \frac{R^2\sigma}{\epsilon_0 z}$$ When \(z\) is inside the sphere, \(z-R < 0\), so \(\sqrt{(R-z)^2}=R-z\) $$V(z) = \frac{R\sigma}{2\epsilon_0 z}\left[(R+z)-(R-z)\right] = \frac{R\sigma}{\epsilon_0}$$
A conical surface (an empty ice-cream cone) carries a uniform surface charge \(\sigma\). The height of the cone is \(h\), as is the radius of the top. Find the potential difference between points \(a\) (the vertex) and \(b\) (the center of the top)
\(r^2 + r^2 = d'^2\), so \(r=d'/\sqrt{2}\) $$V(\vec{a}) = \frac{1}{4\pi\epsilon_0}\int^{\sqrt{2}h}_0\left(\frac{\sigma2\pi r}{d'}\right) dd' = \frac{2\pi\sigma}{4\pi\epsilon_0}\frac{1}{\sqrt{2}}(\sqrt{2} h)=\frac{\sigma h}{2\epsilon_0}$$ $$d = \sqrt{h^2+d'^2-\sqrt{2}hd'}$$ \begin{align} V(\vec{b}) &= \frac{1}{4\pi\epsilon_0}\int^{\sqrt{2}h}_0\left(\frac{\sigma2\pi r}{d}\right)dd'\\ &= \frac{2\pi\sigma}{4\pi\epsilon_0}\frac{1}{\sqrt{2}} \int^{\sqrt{2}h}_0\frac{d'}{\sqrt{h^2+d'^2-\sqrt{2}hd'}}dd'\\ &= \frac{\sigma}{s\sqrt{2}\epsilon_0}\left[\sqrt{h^2+d'^2-\sqrt{2}hd'}+\frac{h}{\sqrt{2}}\ln(2\sqrt{h^2+d'^2-\sqrt{2}hd'}+2d'-\sqrt{2}h)\right]^{\sqrt{2}h}_0\\ &= \frac{\sigma}{s\sqrt{2}\epsilon_0}\left[h+\frac{h}{\sqrt{2}}\ln(2h+2\sqrt{2}h-\sqrt{2}h)-h-\frac{h}{\sqrt{2}}\ln(2h-\sqrt{2}h)\right]\\ &= \frac{\sigma}{2\sqrt{2}\epsilon_0}\frac{h}{2}\left[\ln(2h+\sqrt{2}h)-\ln(2h-\sqrt{2}h)\right]\\ &= \frac{\sigma h}{4\epsilon_0}\ln\left(\frac{2+\sqrt{2}}{2-\sqrt{2}}\right)\\ &= \frac{\sigma h}{4\epsilon_0}\ln\left(\frac{(2+\sqrt{2})^2}{2}\right)\\ &= \frac{\sigma h}{2\epsilon_0}\ln(1+\sqrt{2}) \end{align} So, $$V(\vec{a})-V(\vec{b})=\frac{\sigma h}{2\epsilon_0}\left[1-\ln(1+\sqrt{2})\right]$$
Energy of Electrostatic
The electrical energy of a system is the work done to create the charge distribution.We take infinity as the reference point. Consider a point charge \(q_1\). Putting the point charge requires no work done as no electric field is in the system.
Then put a second point charge \(q_2\). The work done will be $$W_2 = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r_{12}}$$ For a third point charge \(q_3\), the work done will be $$W_2 = \frac{1}{4\pi\epsilon_0}(\frac{q_1q_3}{r_{13}} + \frac{q_2q_3}{r_{23}})$$ Similarly for other point charges coming later. As each charge requires its charge times all the previous charge potential, the total work done for \(n\) charges would be $$W = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^{n} \sum_{j=1}^{i-1} \frac{q_iq_j}{r_{ij}}$$ or equivalently, one may think as a charge is added to the system, the total work done will be sum of each charge potential times the all the late comers $$W = \frac{1}{4\pi\epsilon_0} \sum_{i=1}^{n} \sum_{j=1, j > i}^{n} \frac{q_iq_j}{r_{ij}}$$ So, the total work done could be expressed as \begin{align} W &= \frac{1}{8\pi\epsilon_0} \sum_{i=1}^{n} \sum_{j=1, j \neq i}^{n} \frac{q_iq_j}{r_{ij}}\\ &= \frac{1}{2}\sum_{i=1}^{n}q_i(\sum_{j=1, j\neq i}\frac{1}{4\pi\epsilon_0}\frac{q_j}{r_{ij}}) \\ \end{align} Let \(\textbf{r}_i\) be the position of charge \(q_i\), \(V(r_i)\) be the potential at \(\textbf{r}_i\) by all the charges distribution (not the potential by \(q_i\)). Then the total work done will be $$ \bbox[5px,border:2px solid #666] {W = \frac{1}{2}\sum_{i=1}^{n} q_iV(\textbf{r}_i)}$$ or for continuous charge distribution, $$ \bbox[5px,border:2px solid #666] {W = \frac{1}{2}\int Vdq}$$ \begin{align} W &= \frac{1}{2}\int Vdq \\ &= \frac{1}{2}\int \rho V d\tau \\ &= \frac{\epsilon}{2} \int \nabla \cdot \textbf{E} V d \tau \\ &= \frac{\epsilon}{2} (-\int \textbf{E} \cdot \nabla V d \tau + \oint V \textbf{E} \cdot d\textbf{a} )\\ &= \frac{\epsilon}{2} (\int E^2 d \tau + \oint V \textbf{E} \cdot d\textbf{a} )\\ \end{align} If we integrate over all space, the enclosed surface can be ignored, we will have $$ \bbox[5px,border:2px solid #666] {W = \frac{\epsilon}{2} \int E^2 d \tau }$$
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