Exercises — Electric field of point charge, dipole, ring, disk, line charge (Gauss's law)

Read straight off the parent's summary table.

• (a) Point charge: field lines spread over a sphere of area $4\pi r^2$, so density $\propto 1/r^2$ → $n=2$.
• (b) Infinite line: lines spread over a cylinder of area $2\pi rL$, so $\propto 1/r$ → $n=1$.
• (c) Infinite sheet: lines stay parallel, never spread → $n=0$ (constant field).
• (d) Dipole: the two opposite $1/r^2$ terms nearly cancel, leaving $n=3$. Mnemonic: Point Two, Dipole Three, Line One, Sheet None.

• Ring: no Gaussian surface exists on which $E$ is both constant and perpendicular everywhere (the ring's symmetry is not spherical, cylindrical, or planar). → integrate over elements $dq$.
• Infinite cylinder: full cylindrical symmetry → wrap a coaxial cylinder, $E$ leaves the flux integral → Gauss's law. The deciding question is always: "Can I draw a surface where $\vec{E}\cdot d\vec{A}$ becomes just $E\times A$?" If yes → Gauss. If no → integrate.

WHAT: plug into $E=kq/r^2$. WHY this formula: a single point charge, radial symmetry. $E=\frac{(8.99\times 10^9)(2.0\times 10^{-9})}{(0.30{)}^2}=\frac{17.98}{0.09}\approx 2.0\times 10^2\text{N/C}.$ $E\approx 200\text{N/C}$, pointing radially outward (since $q>0$).

WHY only the axial component survives: for every element $dq$, the element diametrically opposite cancels its sideways pull; only the along-axis piece (weighted by $\cos \theta =x/\sqrt{R^2+x^2}$) is left — see the figure.

WHY Gauss: cylindrical symmetry. From the parent, $E=\frac{2k\lambda }{r}$. $E=\frac{2(8.99\times 10^9)(3.0\times 10^{-6})}{0.20}=\frac{5.394\times 10^4}{0.20}\approx 2.70\times 10^5\text{N/C},$ pointing radially away from the wire.

WHAT: maximise $E(x)=kQx(R^2+x^2{)}^{-3/2}$. WHY differentiate: the peak is where the slope $dE/dx=0$. $\frac{dE}{dx}=kQ\left[(R^2+x^2{)}^{-3/2}+x\cdot (-\frac{3}{2})(R^2+x^2{)}^{-5/2}(2x)\right].$ Factor out $(R^2+x^2{)}^{-5/2}$: $\frac{dE}{dx}=kQ(R^2+x^2{)}^{-5/2}[(R^2+x^2)-3x^2]=kQ(R^2+x^2{)}^{-5/2}(R^2-2x^2).$ Set the bracket to zero: $R^2-2x^2=0\Rightarrow x=R/\sqrt{2}$. ✓ Peak value: with $x^2=R^2/2$, $R^2+x^2=\frac{3}{2}{R}^2$: $E_{max}=\frac{kQ(R/\sqrt{2})}{(\frac{3}{2}{R}^2{)}^{3/2}}=\frac{kQ}{R^2}\cdot \frac{1/\sqrt{2}}{(3/2{)}^{3/2}}=\frac{2kQ}{3\sqrt{3}{R}^2}\approx \frac{0.385kQ}{R^2}.$

(a) $x\ll R$: then $\frac{x}{\sqrt{R^2+x^2}}\to 0$, bracket $\to 1$: $E\to \frac{\sigma }{2{\epsilon }_0}(\text{constant — the infinite sheet}).$ (b) $x\gg R$: write $\frac{x}{\sqrt{R^2+x^2}}=\frac{1}{\sqrt{1+R^2/x^2}}\approx 1-\frac{R^2}{2x^2}$ (binomial, since $R^2/x^2$ is tiny). Bracket $=1-\left(1-\frac{R^2}{2x^2}\right)=\frac{R^2}{2x^2}$. So $E\to \frac{\sigma }{2{\epsilon }_0}\cdot \frac{R^2}{2x^2}=\frac{\sigma \pi R^2}{4\pi {\epsilon }_0{x}^2}=\frac{kQ}{x^2},$ using total charge $Q=\sigma \pi R^2$. A far-away disk looks like a point charge. ✓

From the parent: $E_{ax}=\frac{2kp}{r^3}$ (along $+\vec{p}$) and $E_{eq}=\frac{kp}{r^3}$ (along $-\vec{p}$). $\frac{E_{ax}}{E_{eq}}=\frac{2kp/r^3}{kp/r^3}=2.$ Directions: axial field points along $\vec{p}$ (from $-$ toward $+$ and beyond); equatorial field points opposite to $\vec{p}$. See Electric dipole in a uniform field for how such a dipole then behaves in an external field.

WHY superposition: total field = vector sum of each sheet's field, each of magnitude $\frac{\sigma }{2{\epsilon }_0}$ pointing away from $+$ and toward $-$.

• Between: both fields point the same way (from $+$ plate to $-$ plate), so they add: $E=\frac{\sigma }{2{\epsilon }_0}+\frac{\sigma }{2{\epsilon }_0}=\frac{\sigma }{{\epsilon }_0}.$
• Outside: the two fields point opposite ways and cancel: $E=0$. This is exactly the Parallel plate capacitor result: uniform $\sigma /{\epsilon }_0$ inside, zero outside.

WHAT: at the centre, every element is at the same distance $R$, so each contributes $dE=\frac{kdq}{R^2}$ pointing from the element toward the centre (for $Q>0$, actually away — let's track direction carefully). Set up: the semicircle spans angle $\varphi$ from $-90^{\circ }$ to $+90^{\circ }$ opening toward $+y$. Element charge $dq=\frac{Q}{\pi R}Rd\varphi =\frac{Q}{\pi }d\varphi$ (since linear density $\lambda =Q/(\pi R)$, arc length $Rd\varphi$). By symmetry the $x$-components cancel; only the component along the symmetry axis survives. Each element's field at centre points radially inward-to-outward along its own radius; the surviving component is $dE\cos \varphi$ measured from the axis. $E={\int }_{-\pi /2}^{\pi /2}\frac{kdq}{R^2}\cos \varphi =\frac{k}{R^2}\cdot \frac{Q}{\pi }{\int }_{-\pi /2}^{\pi /2}\cos \varphi d\varphi =\frac{kQ}{\pi R^2}[\sin \varphi {]}_{-\pi /2}^{\pi /2}.$ $E=\frac{kQ}{\pi R^2}(1-(-1))=\frac{2kQ}{\pi R^2}.$ Direction: along the symmetry axis, pointing away from the charged arc (for $Q>0$).

(a) $\frac{\sigma }{2{\epsilon }_0}=\frac{4.0\times 10^{-6}}{2(8.85\times 10^{-12})}=2.26\times 10^5\text{N/C}$. Bracket: $1-\frac{0.20}{\sqrt{0.20^2+0.20^2}}=1-\frac{0.20}{0.2828}=1-0.7071=0.2929$. $E=2.26\times 10^5\times 0.2929\approx 6.62\times 10^4\text{N/C}.$ (b) The infinite sheet would give $2.26\times 10^5\text{N/C}$. At $x=R$ we get only $\approx 29\%$ of that. Why: at a distance comparable to $R$, the disk no longer "fills your sky" — you see its edge, so less charge subtends you, and the field is much weaker than the idealised infinite sheet.

Here $d=2.0\text{mm}=0.002\text{m}$, $p=qd=(1.0\times 10^{-9})(0.002)=2.0\times 10^{-12}\text{C⋅m}$. Direct Coulomb at $r=0.50$ on the axis: the $+q$ is at distance $r-d/2=0.499$, the $-q$ at $r+d/2=0.501$. $E=kq\left[\frac{1}{0.499^2}-\frac{1}{0.501^2}\right]=(8.99)(1.0\times 10^{-9}\cdot 10^9)\text{scaled}.$ Compute carefully: $kq=(8.99\times 10^9)(1.0\times 10^{-9})=8.99$. $E=8.99\left[\frac{1}{0.249001}-\frac{1}{0.251001}\right]=8.99[4.01605-3.98405]=8.99(0.03200)\approx 0.2877\text{N/C}.$ Formula check: $\frac{2kp}{r^3}=\frac{2(8.99\times 10^9)(2.0\times 10^{-12})}{0.50^3}=\frac{0.03596}{0.125}\approx 0.2877\text{N/C}$. ✓ They agree.

WHY $E_x=-dV/dx$: the field is (minus) how steeply potential drops in space — the "downhill slope" of $V$. $V(x)=kQ(R^2+x^2{)}^{-1/2}.$ $\frac{dV}{dx}=kQ\cdot (-\frac{1}{2})(R^2+x^2{)}^{-3/2}(2x)=-\frac{kQx}{(R^2+x^2{)}^{3/2}}.$ $E_x=-\frac{dV}{dx}=\frac{kQx}{(R^2+x^2{)}^{3/2}}.✓$ Exactly the parent's ring formula — two roads, one field.

Which source's field does NOT depend on distance? ::: The infinite sheet, $E=\sigma /2{\epsilon }_0$. Where on a ring's axis is $E$ largest? ::: At $x=R/\sqrt{2}$. Ratio of axial to equatorial dipole field at equal $r$? ::: $2:1$. Field between two opposite sheets of density $\sigma$? ::: $\sigma /{\epsilon }_0$ (outside: $0$). Relation between field and potential on an axis? ::: $E_x=-dV/dx$.