Choose the correct relation between polarisation 'P' and electric susceptibility $^{4} \chi_{e}{ }^{\prime}$…
- $\mathrm{P}=\frac{\chi_{\mathrm{e}}}{\mathrm{E}^{2}}$
- $\mathrm{P}=\frac{\chi_{\mathrm{e}}}{\mathrm{E}}$
- $\mathrm{P}=\chi_{\mathrm{e}} \mathrm{E}$
- $\mathrm{P}=\chi_{\mathrm{e}}^{2} \mathrm{E}$
Solution
Asked in: MHT CET 2020 (20 Oct Shift 2)