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Question

Physics Question on Electric charges and fields

A system has two charges qA=2.5×107Cq_A = 2.5 \times 10^{-7}C and qB=2.5×107Cq_B = -2.5 \times 10^{-7}C located at points A:(0,0,15cm)A : (0, 0,- 15 \,cm) and B:(0,0+15cm)B : (0, 0 + 15 \,cm) , respectively. What is the magnitude and direction of electric dipole moment of the system ?

A

7.5×108cm7.5 \times 10^{-8}\, cm , from positive to negative charge

B

0.75×108cm0.75 \times 10^{-8}\, cm , from negative to positive charge

C

7.5×108cm7.5 \times 10^{-8}\, cm , from negative to positive charge

D

0.75×108cm0.75 \times 10^{-8}\, cm , cm, from positive to negative charge

Answer

7.5×108cm7.5 \times 10^{-8}\, cm , from negative to positive charge

Explanation

Solution

qA=2.5×107Cq_{A}=2.5 \times 10^{-7}C
qB=2.5×107Cq_{B}=-2.5 \times 10^{-7}C

Total charge o f the system, q=qA+qB=0q=q_{A}+q_{B}=0
Distance between two charges at points AA and BB,
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×dp=q_{A}\times d=q_{B}\times d
=2.5×107×0.3=2.5 \times 10^{-7} \times 0.3
=7.5×108cm=7.5\times 10^{-8}\,cm, along positive zz-axis