Q.34 The electrostatic interaction energy between a positively charged atom A and negatively
charged atom B separated by 3 Å in water is -6 kJ/mol. Considering the relative
permittivity of water to be 80, the electrostatic interaction energy in kJ/mol (rounded off to
one decimal place) between atoms A and B in vacuum is _____.
dielectric constant (relative permittivity) of the surrounding medium.
In this article, we calculate how the electrostatic energy changes when moving from
water to vacuum.
Key Concept
Electrostatic interactions are stronger in vacuum because there is no
high-permittivity medium like water to screen the charges.
Electrostatic Energy Formula
According to Coulomb’s law, the electrostatic interaction energy (U) between two
charges is given by:
U = (1 / 4πϵ0ϵr) × (qAqB / r)
- ϵr = relative permittivity
- ϵr = 80 for water
- ϵr = 1 for vacuum
- r = 3 Å
- qA > 0, qB < 0 (attractive interaction)
Given Data
Electrostatic interaction energy in water:
Uwater = −6 kJ/mol
Calculation: Water to Vacuum
Electrostatic energy is inversely proportional to the relative permittivity
(ϵr). Therefore:
Uvacuum = Uwater × ϵr
Substituting values:
Uvacuum = −6 × 80 = −480 kJ/mol
Final Answer
Electrostatic interaction energy in vacuum = −480.0 kJ/mol
Why Is the Energy Stronger in Vacuum?
Water has a high dielectric constant (≈80), which significantly reduces electrostatic
attraction by screening charges. In vacuum (ϵr = 1), no such screening occurs,
resulting in an interaction energy that is 80 times stronger.
Exam Relevance
This type of numerical problem is common in GATE (XL – Biochemistry),
physical chemistry, and biophysics exams. Such questions are often
numerical answer type with no MCQ options.
Key Takeaway
Electrostatic interaction energy scales directly with the inverse of the dielectric
constant. Lower permittivity means stronger (more negative) interaction energy.
