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The unit of polarizability of a molecule is ______.

Asked in GUJCET 2018 · Polarisation and bound charge

(1) C² m¹ N⁻¹
(2) C⁻² m⁻¹ N¹
(3) C⁻² m¹ N⁻¹
(4) C² m⁻¹ N⁻¹

Answer: (1) C² m¹ N⁻¹

Step-by-step solution

Idea: a molecule in a field E acquires an induced dipole moment proportional to the field, p = α E; the constant α is its polarizability.

So α = p/E.

Units: p in C m and E in N C⁻¹.

[α] = (C m)/(N C⁻¹) = C² m¹ N⁻¹.

Why the other options are wrong

(2) This is the unit of E/p, the definition turned upside down. Polarizability is α = p/E.
(3) No single slip gives this; it flips only the sign of the charge exponent. Check: p/E = (C m)/(N C⁻¹), so charge appears squared in the numerator.
(4) No single slip gives this; it flips only the sign of the metre exponent. The dipole moment is charge times length (C m), so the metre appears to the power +1.

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