Force of attraction between two co axial short electric dipoles
Sudharshan Jayakumar , 11 Years ago
Grade 12
3 Answers
Biswajit Das
Last Activity: 9 Years ago
We know that, Coulomb force, F = Kq1q2/d2 Dipole moment, p =qd. So, q1= p1/d and q2= p2/d Putting these values in Coulomb force, we get, F = K(p1/d) (p2/d)/d2 = K (p1) (p2)/d4 Therefore, Force is varies fourth power of distance.
Ajeet Tiwari
Last Activity: 4 Years ago
hello students
We know that, Coulomb force, F = Kq1q2/d2 Dipole moment, p =qd. So, q1= p1/d and q2= p2/d Putting these values in Coulomb force, we get, F = K(p1/d) (p2/d)/d^2 = K (p1) (p2)/d^4 Therefore, Force is varies fourth power of distance.
Hope it helps Thankyou
Yash Chourasiya
Last Activity: 4 Years ago
Dear Students
We know that, Coulomb force, F = Kq1q2/d2 Dipole moment, p = qd. So, q1= p1/d and q2 = p2/d Putting these values in Coulomb force, we get, F = K(p1/d) (p2/d)/d2 = K (p1) (p2)/d4 Therefore, It is clearly seen that Force varies with fourth power of distance.
I hope this answer will help you. Thanks & Regards Yash Chourasiya
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