Superposed D and E Fields of an Infinite Line Charge and a Point Charge: 17-Phys-A3 Electromagnetics, Undated paper, Q7

Question 7: Superposed D and E Fields of an Infinite Line Charge and a Point Charge (20 marks)

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.

Given data
Quantity	Symbol	Value
Line charge (along $z$-axis)	$\rho_l$	10 nC/m
Point charge	$Q_1$	200 nC at $\mathbf r_1=(5,3,2)$ m
Field point	$P$	$(3,4,0)$ m

Find. $\mathbf D$ at $P$ from the line charge alone; the total $\mathbf D$ and $\mathbf E$ at $P$ from both sources.

Approach. An infinite line charge along $z$ gives a purely radial (cylindrical) $\mathbf D=\rho_l/(2\pi s)\,\hat a_s$ by Gauss's law, where $s=\sqrt{x^2+y^2}$ is the perpendicular distance from the $z$-axis (independent of $P$'s $z$-coordinate); a point charge gives the usual $\mathbf D=Q/(4\pi R^2)\hat a_R$ with $\mathbf R$ from the charge to $P$. Superpose, then divide by $\varepsilon_0$ for $\mathbf E$.

Part (a) — D from the line charge at $P=(3,4,0)$. The perpendicular distance from the $z$-axis is $s=\sqrt{3^2+4^2}=5$ m, direction $\hat a_s=(3\mathbf a_x+4\mathbf a_y)/5$: $$\mathbf D_{line}=\frac{\rho_l}{2\pi s}\hat a_s=\frac{10\times10^{-9}}{2\pi(5)}\cdot\frac{3\mathbf a_x+4\mathbf a_y}{5}=\boxed{(0.191\mathbf a_x+0.255\mathbf a_y)\times10^{-9}\ \text{C/m}^2.}$$
Part (b) — D from the point charge at $P$. $\mathbf R=\mathbf r_P-\mathbf r_1=(3-5,4-3,0-2)=(-2,1,-2)$ m, $R=\sqrt{4+1+4}=3$ m, $\hat a_R=(-2,1,-2)/3$: $$\mathbf D_{pt}=\frac{Q_1}{4\pi R^2}\hat a_R=\frac{200\times10^{-9}}{4\pi(9)}\cdot\frac{(-2,1,-2)}{3}$$ $$\mathbf D_{pt}=(-1.179\mathbf a_x+0.589\mathbf a_y-1.179\mathbf a_z)\times10^{-9}\ \text{C/m}^2.$$
Total D and E at P. $$\mathbf D_{tot}=\mathbf D_{line}+\mathbf D_{pt}$$ $$\boxed{\mathbf D_{tot}=(-0.988\mathbf a_x+0.844\mathbf a_y-1.179\mathbf a_z)\times10^{-9}\ \text{C/m}^2}\ (|\mathbf D_{tot}|=1.755\ \text{nC/m}^2)$$ $$\mathbf E_{tot}=\frac{\mathbf D_{tot}}{\varepsilon_0}$$ $$\boxed{\mathbf E_{tot}=-111.6\,\mathbf a_x+95.3\,\mathbf a_y-133.1\,\mathbf a_z\ \text{V/m}}\ (|\mathbf E_{tot}|=198.2\ \text{V/m})$$

Final results
Quantity	Value
$\mathbf D_{line}$ at $P$	$(0.191,\,0.255,\,0)\times10^{-9}$ C/m²
$\mathbf D_{tot}$ at $P$	$(-0.988,\,0.844,\,-1.179)\times10^{-9}$ C/m²
$\mathbf E_{tot}$ at $P$	$(-111.6,\,95.3,\,-133.1)$ V/m, $\|\mathbf E_{tot}\|=198.2$ V/m

Topic: Superposition of Gauss's-law fields from an infinite line charge and a point charge.

Key relations: $\mathbf D_{line}=\rho_l/(2\pi s)\hat a_s$ ($s$ = perpendicular distance from the line); $\mathbf D_{pt}=Q/(4\pi R^2)\hat a_R$; $\mathbf E=\mathbf D/\varepsilon_0$ in free space.

Why this works: Both source fields are found independently from Gauss's law using each one's own symmetry, then simply added as vectors (superposition holds because Maxwell's equations are linear) — there is no interaction term between the two sources' fields.

Common pitfall: Using the full 3-D distance from the $z$-axis instead of the perpendicular (cylindrical-radial) distance $s=\sqrt{x^2+y^2}$ for the line charge (its field does not depend on $z$ at all), and reversing the direction of $\hat a_R$ (must point FROM the point charge TO the field point).

Source: Sadiku, Elements of Electromagnetics, Ch. 4 (Gauss's law: line charge, point charge, superposition).

Question 7 of 10: Superposed D and E Fields of an Infinite Line Charge and a Point Charge

Question 7: Superposed D and E Fields of an Infinite Line Charge and a Point Charge (20 marks)