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Physics Question on Electric Charges and Coulomb's Law

A system has two charge s qA=2.5×10−7Cq_A = 2.5 × 10^{−7} C and qB=−2.5×10−7Cq_B = −2.5 × 10^{−7 }C located at points A: (0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?

Answer

Both the charges can be located in a coordinate frame of reference as shown in the given figure.

At A, amount of charge,qA=2.5×10−7Cq_A = 2.5 × 10^{−7}C
At B, amount of charge,qB=−2.5×10−7Cq_B = −2.5 × 10^{−7} C
Total charge of the system, q=qA+qB=2.5×107C−2.5×10−7C=0q = q_A + q_B = 2.5 × 10^7 C − 2.5 × 10^{ −7} C = 0
Distance between two charges at points A and B,
d=15+15=30cm=0.3md = 15 + 15 = 30 cm = 0.3 m
Electric dipole moment of the system is given by, p
=qA×d=qB×d=2.5×10−7×0.3= q_A × d = q_B × d = 2.5 × 10^{−7 }× 0.3
=7.5×10−8C m= 7.5 × 10^{−8} C\,m along positive z-axis
Therefore, the electric dipole moment of the system is 7.5×10−8C7.5 × 10^{−8} Cm along positive z−axis.