Discussion
Electric susceptibility relates the polarisation to
Answer: D
Correct answer: the field intensity, through the permittivity of free space
\[ \mathbf{P} = \chi_e\,\epsilon_0\mathbf{E}, \] which is the definition of the electric susceptibility. It connects to the permittivity through \( \epsilon_r = 1 + \chi_e \), so a vacuum has zero susceptibility and unit relative permittivity - and the polarisation is driven by \( \mathbf{E} \), the field actually present in the material.
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