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+### Statement
+
+$8.1.20.$ [Insert the problem statement]
+
+### Solution
+
+
+Part (a): Source A (+I) and sink B (–I)
+
+Positions:
+
+· A at (-d/2, 0, 0)
+· B at (+d/2, 0, 0)
+ Separation d.
+
+Individual fields:
+
+· Source A: \displaystyle \mathbf{J}_A = \frac{I}{4\pi r_A^3}\mathbf{r}_A
+· Sink B (outgoing current): \displaystyle \mathbf{J}_B = -\frac{I}{4\pi r_B^3}\mathbf{r}_B
+
+Plane of symmetry: x = 0, perpendicular to line AB at its midpoint.
+
+By symmetry, on the plane x = 0:
+
+· r_A = r_B = r = \sqrt{(d/2)^2 + y^2 + z^2}
+· The transverse (y, z) components of \mathbf{J}_A and \mathbf{J}_B cancel.
+· The x-components add constructively.
+
+J_x(0, y, z) = \frac{I}{4\pi r^3}\left(\frac{d}{2}\right) + \frac{I}{4\pi r^3}\left(\frac{d}{2}\right) = \frac{I d}{4\pi r^3}
+
+J_y = J_z = 0.
+
+Thus, the volume current density in the symmetry plane is purely perpendicular to the plane:
+
+\boxed{\mathbf{J} = \frac{I d}{4\pi \big[(d/2)^2 + \rho^2\big]^{3/2}}\;\hat{x}}, \qquad \rho = \sqrt{y^2 + z^2}.
+
+Total current through the plane:
+Integrate over the entire plane x = 0:
+
+I_{\text{plane}} = \int_0^\infty J_x(\rho)\, 2\pi\rho\, d\rho
+= \int_0^\infty \frac{I d}{4\pi (\frac{d^2}{4}+\rho^2)^{3/2}}\, 2\pi\rho\, d\rho.
+
+With a = d/2, the integral equals 1/a = 2/d. Then:
+
+I_{\text{plane}} = \frac{I d}{2} \cdot \frac{2}{d} = I.
+
+\boxed{I_{\text{total}} = I}
+
+All the current leaving A crosses the symmetry plane toward B.
+
+If B also carries a current I (both are sources):
+Now \mathbf{J}_B = +\frac{I}{4\pi r_B^3}\mathbf{r}_B. On the plane x = 0 the x-components oppose:
+
+J_{Ax} = \frac{I d}{8\pi r^3},\quad J_{Bx} = -\frac{I d}{8\pi r^3}.
+
+Result: \mathbf{J} = 0 on the plane, total current = 0. No net flow by symmetry.
+
+---
+
+Part (b): Point source at depth h below the Earth's surface
+
+Model: Conducting half-space (z > 0), insulating surface at z = 0 (no current flow into the air). Real source at (0,0,h) with current I.
+
+Method of images:
+To satisfy J_z = 0 at z = 0, an identical image source of current I is added at (0,0,-h), considering the full space.
+
+Current density on the surface (z = 0):
+
+· Vector from real source: \mathbf{r}_1 = (x, y, -h), r_1 = \sqrt{\rho^2 + h^2}.
+· Vector from image source: \mathbf{r}_2 = (x, y, h), r_2 = r_1 = r.
+
+\mathbf{J}_{\text{surf}} = \frac{I}{4\pi r^3}\mathbf{r}_1 + \frac{I}{4\pi r^3}\mathbf{r}_2
+= \frac{I}{4\pi r^3}(2x, 2y, 0)
+= \frac{I}{2\pi (\rho^2 + h^2)^{3/2}}\,(x\hat{x} + y\hat{y}).
+
+It is radially outward from the point directly above the source (\rho = 0). Magnitude:
+
+\boxed{J(\rho) = \frac{I}{2\pi} \frac{\rho}{(\rho^2 + h^2)^{3/2}} }.
+
+The surface current density is purely horizontal and decreases with radial distance \rho.
+
+#### Answer
+
+[Insert a concise answer or boxed result]