The problem in the image reads: **2.14** A homogeneous line charge $\ell_e$ is located on the *z*-axis between $z = -a$ and $z = a$. Calculate components $E_y(y,z)$ and $E_z(y,z)$ of the electric field at point $(x=0, y, z)$.

What concepts are being tested?

1. Coulomb’s Law for a point charge
   $\vec E = \frac{1}{4\pi\varepsilon_0}\,\frac{q\,\hat r}{r^2}$
2. Treating a line charge as many point charges
   $q \longrightarrow \lambda\,dz'$ where $\lambda$ is the constant line charge density.
3. Vector addition via integration – add the infinitesimal fields produced by elements between $z'=-a$ and $z'=+a$.
4. Symmetry arguments – components that cancel out are identified immediately, saving algebra.

Logical chain of thought

1. Choose an element at position $(0,0,z')$ on the rod. Observation point is $(0,y,z)$.
2. Write the displacement vector from the element to the point:
   $\vec R = (0,\,y,\,z-z')$
3. Magnitude of that vector:
   $R = \sqrt{y^2 + (z - z')^2}$
4. Apply Coulomb’s law for the element:
   $d\vec E = \frac{1}{4\pi\varepsilon_0}\,\frac{\lambda\,dz'\,\vec R}{R^3}$
5. Identify components: the $x$-component is zero; keep only $E_y$ and $E_z$ terms.
6. Set up two separate integrals (one for $E_y$, one for $E_z$) with limits $-a$ to $+a$.
7. Use a substitution $u = z - z'$ to simplify both integrals.
8. Recognise standard integral forms:
   • $\int \frac{du}{(u^2 + y^2)^{3/2}}$
   • $\int \frac{u\,du}{(u^2 + y^2)^{3/2}}$
     and recall their antiderivatives.
9. Evaluate between limits to get concise closed-form answers.
10. Check special cases (e.g. $y \to 0$ or $a \to \infty$) to be sure formula makes physical sense.

Every step flows naturally from the previous one, keeping symmetry and calculus hand-in-hand.

Imagine a glowing stick

• Think of a thin light-stick that starts at some point below you and ends the same distance above you on a straight vertical line.
• That stick is packed evenly with electric charge. Because of this, every tiny bit of the stick gives you a little electric push.
• We want to know how strong the push is sideways (along $y$) and straight up-down (along $z$) at a spot in the air that is right in front of the stick (no $x$ distance).
• To get the total push we add up (integrate) the pushes from all the tiny pieces of the stick.
• Symmetry tells us:
  • There is no push in the $x$-direction because the stick sits exactly on the $z$-axis and we are also on the $x=0$ plane.
  • Only the $y$ and $z$ directions matter.
• Finally the maths (integration) gives neat formulas which tell us how the field changes with your distance $y$ from the stick and how high $z$ you are.

That’s the story! 🙂

Step-by-step calculation

Let the uniform line charge density be $\lambda$ ($C/m$). Observation point: $(0,\,y,\,z)$. Source point on line: $(0,\,0,\,z')$ with $z'\in[-a,a]$.

Displacement vector $\vec R = \big(0,\,y,\,z - z'\big)$

Magnitude $R = \sqrt{y^2 + (z - z')^2}$

Infinitesimal field $d\vec E = \frac{1}{4\pi\varepsilon_0}\,\frac{\lambda\,dz'\,\vec R}{R^3}$

Break into components: $dE_y = \frac{1}{4\pi\varepsilon_0}\,\frac{\lambda\,y\,dz'}{\big[y^2 + (z - z')^2\big]^{3/2}}$ $dE_z = \frac{1}{4\pi\varepsilon_0}\,\frac{\lambda\,(z - z')\,dz'}{\big[y^2 + (z - z')^2\big]^{3/2}}$

1. $E_y$

Make substitution $u = z - z'$ $\Rightarrow$ $dz' = -du$. Limits change: $z' = -a \;\Rightarrow\; u = z + a$
$z' = +a \;\Rightarrow\; u = z - a$

$E_y = \frac{1}{4\pi\varepsilon_0}\,\lambda\,y\int_{-a}^{a}\!\frac{dz'}{\big[y^2 + (z - z')^2\big]^{3/2}}$

becomes $E_y = \frac{1}{4\pi\varepsilon_0}\,\lambda\,y\int_{z - a}^{z + a}\!\frac{du}{\big(y^2 + u^2\big)^{3/2}}$

Use the standard integral $\int \frac{du}{(u^2 + y^2)^{3/2}} = \frac{u}{y^2\sqrt{u^2 + y^2}}$

Hence $E_y = \frac{1}{4\pi\varepsilon_0}\,\lambda\,\frac{1}{y}\Bigg[\frac{u}{\sqrt{u^2 + y^2}}\Bigg]_{u = z - a}^{z + a}$

$\boxed{\displaystyle E_y(y,z) = \frac{\lambda}{4\pi\varepsilon_0}\,\frac{1}{y}\left[\frac{z + a}{\sqrt{y^2 + (z + a)^2}} - \frac{z - a}{\sqrt{y^2 + (z - a)^2}}\right]}$

2. $E_z$

Same substitution.

$E_z = \frac{1}{4\pi\varepsilon_0}\,\lambda\int_{z - a}^{z + a}\!\frac{ -u\,du}{\big(y^2 + u^2\big)^{3/2}}$

Use $\int \frac{u\,du}{(u^2 + y^2)^{3/2}} = -\frac{1}{\sqrt{u^2 + y^2}}$

Therefore $E_z = \frac{1}{4\pi\varepsilon_0}\,\lambda\Bigg[\frac{1}{\sqrt{u^2 + y^2}}\Bigg]_{u = z - a}^{z + a}$

$\boxed{\displaystyle E_z(y,z) = \frac{\lambda}{4\pi\varepsilon_0}\left[\frac{1}{\sqrt{y^2 + (z - a)^2}} - \frac{1}{\sqrt{y^2 + (z + a)^2}}\right]}$

Both expressions reduce correctly:

• If $y \to 0$ (point on the axis) we recover the field of a finite line at its own axis.
• If $a \to \infty$ (infinite line) we get $E_y = \dfrac{\lambda}{2\pi\varepsilon_0 y}$ and $E_z \to 0$, the standard result.