Question 7 of 10: Line charge and point charge by Gauss' law and superposition

Question 7: Line charge and point charge by Gauss' law and superposition

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. An infinite line charge of 10 nC/m along the $z$-axis together with a 200 nC point charge at $(5, 3, 2)$ m, both in free space; the field point is $P(3, 4, 0)$ m.

Find. $\mathbf{D}$ at $P$ from the line charge alone by Gauss' law; then the total $\mathbf{D}$ and total $\mathbf{E}$ at $P$ from both sources.

[Figure not reproduced: Figure for Q7 (redrawn). The line charge sets a purely radial D at P; the point charge adds a Coulomb term along R = r_P − r_Q. See the official exam paper.]

Approach. Use a coaxial Gaussian cylinder for the line charge (the only surface on which its $\mathbf{D}$ is uniform and normal), add the point charge by Coulomb's law in vector form, and superpose — both are linear-medium fields, so the flux densities add directly.

1. (a) Derive the line-charge field from Gauss' law. An infinite uniform line on the $z$-axis has a field that can only be radial and can only depend on the perpendicular distance $\rho$. Taking a coaxial cylinder of radius $\rho$ and length $L$, the flat ends contribute nothing because $\vec{D}$ is tangential there, so $$\oint\vec{D}\cdot d\vec{S} = D_\rho(2\pi\rho L) = Q_{enc} = \rho_l L \;\Rightarrow\; \vec{D} = \frac{\rho_l}{2\pi\rho}\hat{a}_\rho$$ The length $L$ cancels, which is why the result is independent of how much of the line is enclosed.
2. Locate the field point relative to the axis. Only the perpendicular distance matters, and the $z$-coordinate of $P$ is irrelevant: $$\rho = \sqrt{3^2 + 4^2} = 5.00\ \text{m}, \qquad \hat{a}_\rho = \frac{3\mathbf{a}_x + 4\mathbf{a}_y}{5} = 0.6\mathbf{a}_x + 0.8\mathbf{a}_y$$ a clean 3–4–5 triangle, so the unit vector is exact.
3. Evaluate the line-charge flux density. $$|\vec{D}_l| = \frac{10\times10^{-9}}{2\pi(5.00)} = 3.183\times10^{-10}\ \text{C/m}^2 = 0.3183\ \text{nC/m}^2$$ and in Cartesian form $$\boxed{\vec{D}_l = 0.1910\,\mathbf{a}_x + 0.2546\,\mathbf{a}_y\ \text{nC/m}^2}$$ with no $z$-component at all. Checking back through Gauss' law, $|\vec{D}_l|(2\pi\rho)(1\ \text{m}) = 10$ nC, which is exactly the charge on 1 m of the line.
4. (b) Add the point charge by Coulomb's law. The separation vector from the charge to the field point is $$\vec{R} = \mathbf{r}_P - \mathbf{r}_1 = (3-5)\mathbf{a}_x + (4-3)\mathbf{a}_y + (0-2)\mathbf{a}_z = -2\mathbf{a}_x + \mathbf{a}_y - 2\mathbf{a}_z$$ so $|\vec{R}| = \sqrt{4 + 1 + 4} = 3.00$ m exactly and $\hat{a}_R = (-2\mathbf{a}_x + \mathbf{a}_y - 2\mathbf{a}_z)/3$. Then $$\vec{D}_Q = \frac{Q_1}{4\pi|\vec{R}|^2}\hat{a}_R = \frac{200\times10^{-9}}{4\pi(9)}\hat{a}_R = 1.768\,\hat{a}_R\ \text{nC/m}^2$$ $$\vec{D}_Q = -1.1789\,\mathbf{a}_x + 0.5895\,\mathbf{a}_y - 1.1789\,\mathbf{a}_z\ \text{nC/m}^2$$ Note the point charge dominates: its 1.768 nC/m$^2$ is five and a half times the line's contribution.
5. Superpose the two flux densities. Free space is linear, so the fields add component by component: $$\boxed{\vec{D}_{tot} = -0.9879\,\mathbf{a}_x + 0.8441\,\mathbf{a}_y - 1.1789\,\mathbf{a}_z\ \text{nC/m}^2}$$ whose magnitude is $|\vec{D}_{tot}| = 1.755$ nC/m$^2$. Because the line's $\mathbf{a}_x$ term is positive and the point charge's is negative, they partly cancel — the total is slightly smaller than the point-charge term alone, which is the kind of partial cancellation a magnitude-only calculation would miss.
6. Convert to the total electric field. In free space $\vec{E} = \vec{D}/\epsilon_0$, so dividing each component by $8.854\times10^{-12}$ F/m: $$\boxed{\vec{E}_{tot} = -111.58\,\mathbf{a}_x + 95.34\,\mathbf{a}_y - 133.15\,\mathbf{a}_z\ \text{V/m}}$$ with magnitude $|\vec{E}_{tot}| = 198.2$ V/m. As a consistency check, $|\vec{D}_{tot}|/\epsilon_0 = 1.7545\times10^{-9}/8.854\times10^{-12} = 198.2$ V/m, matching the vector magnitude.