Question

A system has two changes $q_A=2.5\times {10}^{-7}C$ $and{q}_B=-2.5\times {10}^{-7}C$ located at point $A(0,0,-15)cm$and $B(0,0,15)cm$respectively. What is the total charge and electric dipole moment vector of the system?

Answer 1

Hint
We know that the electric dipole moment is defined as the measure of the separation of the positive and negative electrical charges within a system, that is, a measure of the overall polarity, the dipole is represented by a vector from the negative charge towards the positive charge. Based on this concept we have to solve this question.

Complete step-by step answer:
At A, amount of charge $q_A=2.5\times {10}^{-7}C$(Given)
At B, amount of charge $q_B=-2.5\times {10}^{-7}$
Total charge of the system is $q=q_A+q_B=2.5\times {10}^{-7}C=0$
Distance between two charges at points A and B,
$d=15+15=30cm=0.3m$
Electric dipole moment of the system is given by,
$P=q_A\times d=2.5\times {10}^{-7}\times 0.3=7.5\times {10}^{-8}cm$ along the positive z-axis.
Therefore, the electric dipole moment of the system is $7.5\times {10}^{-8}$ along the positive z-axis.

Note
We should know that a dipole moment gives us an idea of the turning force a fixed charge has on a dipole in a molecule. The moment depends on the charge at the end of the dipole and its distance from the charge at the other end of the dipole. So, we know that the larger the difference in electronegativity, the larger the dipole moment. The distance between the charge separation is also a deciding factor into the size of the dipole moment.