A system has two charges qA = 2.5 × 10−7 C and qB = −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?
According to the figure given below both the charges can be located in a coordinate frame of reference.
At A, amount of charge, qA = 2.5 × 10−7C (Given)
At B, amount of charge, qB = −2.5 × 10−7 C (Given)
Total charge of the system,
q = qA + qB = 2.5 × 10-7 C − 2.5 × 10−7 C = 0
Distance between two charges at points A and B,
d = 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 = 7.5 × 10−8 C m along positive z-axis
Therefore, the electric dipole moment of the system is 7.5 × 10−8 C m along positive z−axis.
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Welcome to the NCERT Solutions for Class 12 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 9: A system has two charges qA = 2.5 × 10−7 C and qB = −2.5 × 10−7 C loca....
Comments
it is simple 15-(-15)=30 cm answer is 0.03 m
how to find distance in graph points?????????????????????????
Exactly chaitanya
The dipole must be along negative z axis