checkpoint 1

concepts/em/u8/e8-4-continuous-distributions.tex

@@ -0,0 +1,156 @@
+\subsection{Fields of Continuous Charge Distributions}
+
+This subsection extends electric-field superposition from point charges to rods, arcs, rings, surfaces, and volumes by replacing discrete sums with integrals over charge elements.
+
+\dfn{Charge densities and differential field contribution}{Let the field point have position vector $\vec{r}$. Let a small source element at position vector $\vec{r}'$ carry charge $dq$. Define
+\[
+\vec{R}=\vec{r}-\vec{r}',
+\qquad
+R=|\vec{R}|,
+\qquad
+\hat{R}=\frac{\vec{R}}{R}.
+\]
+For a continuous distribution, the charge element is written as
+\[
+dq=\lambda\,dl
+\qquad\text{(line charge)},
+\qquad
+dq=\sigma\,dA
+\qquad\text{(surface charge)},
+\qquad
+dq=\rho\,dV
+\qquad\text{(volume charge)},
+\]
+where $\lambda$ is linear charge density in $\mathrm{C/m}$, $\sigma$ is surface charge density in $\mathrm{C/m^2}$, and $\rho$ is volume charge density in $\mathrm{C/m^3}$. The electric-field contribution of the source element at the field point is
+\[
+d\vec{E}=k\frac{dq}{R^2}\hat{R}=k\frac{dq}{R^3}\vec{R},
+\]
+where
+\[
+k=\frac{1}{4\pi\varepsilon_0}=8.99\times 10^9\,\mathrm{N\,m^2/C^2}.
+\]
+}
+
+\thm{Continuous superposition integral for electric field}{For a static continuous charge distribution, the net electric field at the field point $\vec{r}$ is the vector integral
+\[
+\vec{E}(\vec{r})=\int d\vec{E}
+=k\int \frac{1}{R^3}\vec{R}\,dq.
+\]
+Equivalently,
+\[
+\vec{E}(\vec{r})
+=k\int \frac{1}{R^3}\vec{R}\,\lambda\,dl,
+\qquad
+\vec{E}(\vec{r})
+=k\int \frac{1}{R^3}\vec{R}\,\sigma\,dA,
+\qquad
+\vec{E}(\vec{r})
+=k\int \frac{1}{R^3}\vec{R}\,\rho\,dV,
+\]
+depending on the geometry. In practice, write $d\vec{E}$ for one source element, resolve it into components, use symmetry to identify any canceling components, and integrate only the nonzero component(s).}
+
+\ex{Illustrative example}{A uniformly charged semicircular arc of radius $R$ lies above the $x$-axis and is centered at the origin. Let its total charge be $Q>0$. Find the electric field at the center.
+
+The arc length is $\pi R$, so the linear charge density is
+\[
+\lambda=\frac{Q}{\pi R}.
+\]
+Let $\theta$ denote the polar angle of a source element, measured from the positive $x$-axis, with $0\le \theta \le \pi$. Then
+\[
+dq=\lambda R\,d\theta.
+\]
+Each source element is distance $R$ from the center, so
+\[
+dE=k\frac{dq}{R^2}.
+\]
+By symmetry, the $x$-components cancel. The $y$-components all point downward, so
+\[
+dE_y=-dE\sin\theta=-k\frac{dq}{R^2}\sin\theta.
+\]
+Substitute $dq=\lambda R\,d\theta$:
+\[
+dE_y=-k\frac{\lambda}{R}\sin\theta\,d\theta.
+\]
+Integrate from $0$ to $\pi$:
+\[
+E_y=-k\frac{\lambda}{R}\int_0^\pi \sin\theta\,d\theta
+=-k\frac{\lambda}{R}(2).
+\]
+Therefore,
+\[
+\vec{E}=-\frac{2kQ}{\pi R^2}\hat{\jmath}.
+\]
+The field points downward because the positive charges on the upper arc repel a positive test charge at the center.}
+
+\nt{For continuous distributions, the hardest step is usually not the integral but the geometry. Start with $d\vec{E}$ from one source element, then ask which components cancel by symmetry. On a ring, sideways components cancel and only the axial component survives. On a symmetric finite line, horizontal components cancel at the perpendicular bisector and only the perpendicular component survives. Also choose $dq$ to match the object's dimension: use $dq=\lambda\,dl$ for rods and arcs, $dq=\sigma\,dA$ for sheets, and $dq=\rho\,dV$ for three-dimensional charge distributions.}
+
+\qs{Worked AP-style problem}{A thin ring of radius $a$ is centered at the origin and lies in the $yz$-plane. The ring carries total charge $Q$ distributed uniformly around its circumference. Let point $P$ lie on the ring's axis at position
+\[
+\vec{r}_P=x\hat{\imath},
+\]
+where $x>0$. Let
+\[
+k=\frac{1}{4\pi\varepsilon_0}.
+\]
+Find the electric field $\vec{E}$ at point $P$ in terms of $Q$, $a$, $x$, and $k$.}
+
+\sol Let $\lambda$ denote the ring's linear charge density. Since the ring circumference is $2\pi a$,
+\[
+\lambda=\frac{Q}{2\pi a}.
+\]
+
+Choose a small source element $dq$ on the ring. Let $\vec{R}$ denote the displacement vector from that source element to point $P$. Every source element on the ring is the same distance from $P$, so
+\[
+R=\sqrt{x^2+a^2}.
+\]
+The magnitude of the field due to $dq$ is therefore
+\[
+dE=k\frac{dq}{R^2}=k\frac{dq}{x^2+a^2}.
+\]
+
+Now use symmetry. For each source element on the ring, there is an opposite element whose field contribution has the same magnitude. Their components in the $y$- and $z$-directions cancel, while their components along the axis add. Therefore the net field must point along $\hat{\imath}$, so we only need the $x$-component of $d\vec{E}$.
+
+Let $\phi$ denote the angle between $\vec{R}$ and the positive $x$-axis. Then
+\[
+\cos\phi=\frac{x}{R}=\frac{x}{\sqrt{x^2+a^2}}.
+\]
+So the axial component of the differential field is
+\[
+dE_x=dE\cos\phi
+=\left(k\frac{dq}{R^2}\right)\left(\frac{x}{R}\right)
+=k\frac{x\,dq}{R^3}.
+\]
+Since $R=\sqrt{x^2+a^2}$ is constant over the ring,
+\[
+dE_x=k\frac{x\,dq}{(x^2+a^2)^{3/2}}.
+\]
+
+Integrate all the way around the ring:
+\[
+E_x=\int dE_x
+=k\frac{x}{(x^2+a^2)^{3/2}}\int dq.
+\]
+But
+\[
+\int dq=Q,
+\]
+so
+\[
+E_x=k\frac{Qx}{(x^2+a^2)^{3/2}}.
+\]
+
+Therefore,
+\[
+\vec{E}=k\frac{Qx}{(x^2+a^2)^{3/2}}\hat{\imath}.
+\]
+If $Q>0$, the field points in the $+\hat{\imath}$ direction, and if $Q<0$, it points in the $-\hat{\imath}$ direction.
+
+This result also passes two quick checks. If $x=0$, then
+\[
+\vec{E}=\vec{0},
+\]
+which matches the symmetry at the ring's center. If $x\gg a$, then $x^2+a^2\approx x^2$, so
+\[
+\vec{E}\approx k\frac{Q}{x^2}\hat{\imath},
+\]
+which is the field of a point charge $Q$ far away from the ring.