what is meant by ....the electric field due to diapole falls off at large distance faster than like 1/r^2
masratalisha99:
As fas as I know, it falls by the rate of 1/r^3 and not 1/r^2. So maybe you should correct the question too.
nope its given in ncert r^2
Yup I just saw it, dropping by 1/r^3 is a different concept. You have posted it correct. I will try to solve it.
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Okay so I kinda googled about this, and I found various sources where it is given it falls at a rate of 1/r^3, but it's also given, as you mentioned that it falls off faster than 1/r^2 at large distances.
What I have understood from this, that their field decreases by the inverse of square of the distance between them. It is simple to assume that when the two charges that constitute the dipole are kept at larger disrances, than they cannot cancel out each other completely. There's a partial cancellation, which is obviously direction dependant. And most probably, due to this partial cancellation, the electric fields falls of, because as I said earlier, the whole field strength is very much direction dependant.
Hope your doubt got clear.
What I have understood from this, that their field decreases by the inverse of square of the distance between them. It is simple to assume that when the two charges that constitute the dipole are kept at larger disrances, than they cannot cancel out each other completely. There's a partial cancellation, which is obviously direction dependant. And most probably, due to this partial cancellation, the electric fields falls of, because as I said earlier, the whole field strength is very much direction dependant.
Hope your doubt got clear.
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