Question

An electrically neutral molecule is collinear with (and located between) two charged particles, one carrying a charge of +3.56 μC and the other carrying a charge of -1.05 μC. The center of the molecule is 2.57 μm from each particle. Part A If the vector sum of the electric forces exerted on the molecule is 45.0 nN , what is the polarizability of the molecule? Express your answer using two significant digits. α =

Answer 1

Dipole Moment of the molecule about the center point will be

$p= q_1a+q_2a= (3.56 \times 10^{-6})(2.57 \times 10^{-6})+(-1.05 \times 10^{-6})(2.57 \times 10^{-6})$

$p= 6.45 \times 10^{-12} C.m$ ....................(1)

We know that Electric Field emerges from positive charge and merge into negative charge. Thus at the center point, Electric Field due to both charges will point in one direction, i.e. towards negative charge. Then we can write net Electric Field at the center will be

   $E_n= |E_1|+|E_2|= \frac{|q_1|}{4 \pi \epsilon_o a^2}+\frac{|q_2|}{4 \pi \epsilon_o a^2}$

   $E_n= \frac{(|q_1|+|q_2|)}{4 \pi \epsilon_o a^2}$

using given values in above,

   $E_n= \frac{(|3.56 \times 10^{-6}|+|(-1.05 \times 10^{-6})|)}{4 \times 3.14 \times 8.85 \times 10^{-12} \times (2.57 \times 10^{-6})^2}$

   $E_n= 6.279 \times 10^{9} N/C$

Then polarizability of the molecule is given by

$\alpha= \frac{p}{E_n}$

$\alpha= \frac{6.45 \times 10^{-12}}{6.279 \times 10^{9}}= 1.027 \times 10^{-21}C^2.m/N$