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(B) The volume of the cube is $V = (1 \ cm)^3 = 10^{-6} \ m^3$.The number density of molecules is $n = \frac{100}{10^{-6}} = 10^8 \ m^{-3}$.The polari

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$5$

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(B) The volume of the cube is $V = (1 \ cm)^3 = 10^{-6} \ m^3$.The number density of molecules is $n = \frac{100}{10^{-6}} = 10^8 \ m^{-3}$.The polarization $P$ is given by $P = n \cdot p$,where $p$ is the induced dipole moment.$P = 10^8 	imes 0.2 	imes 10^{-6} = 0.02 \ C \cdot m^{-2}$.The relation between polarization $P$,electric susceptibility $\chi_e$,and electric field $E$ is $P = \epsilon_0 \chi_e E$. However,in many contexts,the susceptibility $\chi_e$ is defined via $P = \chi_e E$.Using $P = \chi_e E$,we get $\chi_e = \frac{P}{E} = \frac{0.02}{4} = 0.005$.Re-evaluating based on the provided solution logic: $\chi_e = \frac{n \cdot p}{E} = \frac{10^8 	imes 0.2 	imes 10^{-6}}{4} = \frac{20}{4} = 5$.

$A$ cube of side $1 \ cm$ contains $100$ molecules,each having an induced dipole moment of $0.2 \times 10^{-6} \ C \cdot m$ in an external electric field of $4 \ N \cdot C^{-1}$. The electric susceptibility of the material is .... $C^2 \cdot N^{-1} \cdot m^{-2}$.

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