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Asked in GUJCET 2020 · Polarisation and bound charge
The textbook defines electric susceptibility through ⃗P=χₑ⃗E, so χₑ carries the same unit as ε₀, C² N⁻¹ m⁻².
In a slab of dielectric between charged plates the bound surface charge is σₚ=P=χₑE, so ε₀E=σ-χₑE, giving E=σ/(ε₀+χₑ).
Comparing with E=σ/(ε₀εᵣ): ε₀εᵣ=ε₀+χₑ, so χₑ=ε₀(εᵣ-1).
χₑ=8.85×10⁻¹²×(80-1)=6.99×10⁻¹⁰.
So χₑ≈7×10⁻¹⁰ C² N⁻¹ m⁻². (Were susceptibility defined as a pure number instead, it would be εᵣ-1=79; keep to the textbook's definition.)
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