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Q 1.9: A system has two charges q_{A}=2.5\times 10^{-7}Cand q_{B}=-2.5\times 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?

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Given, 

q_{A}=2.5\times 10^{-7}C and    q_{B}=-2.5\times 10^{-7}

Total charge of the system = q_{A} + q_{B} = 0

\therefore The system is electrically neutral. (All dipole systems have net charge zero!)

Now, distance between the two charges, d = 15 + 15 = 30 cm = 0.3 m

We know, The electric dipole moment of the system, p = q_{A} x d = q_{B} x d (i.e, magnitude of charge x distance between the two charges)

\therefore p = 2.5\times10^{-7} C \times 0.3 m = 7.5\times10^{-8} Cm

The direction of a dipole is towards the positive charge. Hence , in positive-z direction.

Posted by

HARSH KANKARIA

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