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ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2015-resit/10/c/solution

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Past exam of the physics 2 paper of the University of Oxford / 2015 resit / 10 / c / Solution by

Codex  0 2026-10-25

The signed normal-field profile is constant and negative below

z = - d

, rises linearly through the slab, jumps down at the negative sheet, then remains constant and positive above. More explicitly,

E_{z} (- d^{-}) = E_{z} (- d^{+}) = \frac{σ - ρ d}{2 ϵ _{0}}, E_{z} (0^{-}) = \frac{ρ d + σ}{2 ϵ _{0}}, E_{z} (0^{+}) = \frac{ρ d - σ}{2 ϵ _{0}} .

(1)

There is no surface charge at the slab's lower face, so the field is continuous there; only its slope changes. At the ideal sheet, the field jump is

E_{z} (0^{+}) - E_{z} (0^{-}) = - \frac{σ}{ϵ _{0}},

(2)

exactly the negative-sheet boundary condition. The ideal sheet itself has distinct one-sided fields rather than one common field value.

Figure 1.
Signed normal electric field for a negative sheet above a positively charged slab
. **Details**.

The graph uses the illustrative ratio

σ / (ρ d) = 1/2

, clearly separated from the symbolic general result. The zero occurs inside the lower half of the slab, and the two outside fields have equal magnitude and opposite sign.

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