Coaxial cable potential and a misleading multimeter
A coaxial cable 1.20 m long has a solid inner conductor of radius 0.80 mm and a coaxial outer conductor whose inner radius is 4.00 mm. The space between them is evacuated. The inner conductor carries +3.5 nC and the outer conductor carries −3.5 nC, uniformly distributed along the length. A student models the field as that of an infinite line, calculates the potential difference, and then measures it with a digital multimeter whose input resistance is 10 MΩ. The meter briefly reads 42 V and then falls to zero. The student concludes that the infinite-line model is wrong by a factor of two. The lab supervisor suspects the measurement method rather than the model. The charge was placed on the cable with a triboelectric rod, not maintained by a battery. Use ε0=8.85×10^-12 F m^-1.
1. Calculate the linear charge density, the electric field at r=2.0 mm, and the predicted potential difference between the conductors. Compare with the 42 V reading.
Read the worked answer
Linear charge density: λ = Q/L = 3.5×10^-9/1.20 = 2.92×10^-9 C/m.
For an infinite line, E(r) = λ/(2π ε0 r). At r = 2.0×10^-3 m, the denominator is 2π(8.85×10^-12)(2.0×10^-3) = 1.11×10^-13, so E = 2.92×10^-9/1.11×10^-13 ≈ 2.6×10^4 V/m, directed radially outward.
The potential difference is V = ∫E dr from a to b = [λ/(2π ε0)] ln(b/a) = 2kλ ln(b/a). Here 2kλ = 2(8.99×10^9)(2.92×10^-9) ≈ 52.5 V, and ln(4.00/0.80) = ln 5 ≈ 1.61, giving V ≈ 84.5 V.
The predicted 84.5 V is about twice the 42 V reading, but the reading is not an equilibrium value. A factor-of-two discrepancy alone does not prove the model wrong; the measurement must be checked first.
2. Explain why the multimeter reading decays to zero. Estimate the RC time constant and decide whether the infinite-line model is invalidated by the 42 V reading.
Read the worked answer
The cable is a cylindrical capacitor. Its capacitance is C = 2π ε0 L/ln(b/a) = (5.56×10^-11×1.20)/1.61 ≈ 4.15×10^-11 F = 41.5 pF. The multimeter's 10 MΩ input resistance forms a discharge path: τ = RC = (10×10^6)(4.15×10^-11) ≈ 4.15×10^-4 s, less than a millisecond. The charge placed by the triboelectric rod is small and is not replenished, so the meter discharges the cable almost immediately. The displayed 42 V is a transient sample, not a static potential difference.
Therefore the 42 V reading does not invalidate the infinite-line model. To test the model, use an electrometer with input resistance of order 10^14 Ω or more, or a non-contact electrostatic probe, or maintain the charge with a battery. The infinite-line approximation is also reasonable in the middle of a 1.20 m cable whose outer radius is only 4 mm; end effects are confined to a few radii near the ends.
3. If a dielectric with κ=2.3 fills the space between the conductors, how do E, V, and C change when the charge is fixed? What changes if the cable is instead connected to a battery?
Read the worked answer
With fixed free charge Q, inserting the dielectric reduces the electric field by the factor κ: at r = 2.0 mm, E becomes 2.6×10^4/2.3 ≈ 1.1×10^4 V/m. The potential difference is also reduced by κ: V ≈ 84.5/2.3 ≈ 37 V. The capacitance increases by κ: C ≈ 2.3×41.5 ≈ 95 pF. The energy stored, Q^2/(2C), decreases, with the difference going into the dielectric polarization and mechanical insertion work.
If the cable is connected to a battery, the potential difference V is fixed instead. Then E at a given r is fixed by V and the geometry, E = V/[r ln(b/a)], so E does not change. The free charge on the conductors increases by κ because C increases, and the stored energy increases. The dielectric polarization increases the displacement field D while the electric field E remains set by the battery voltage.