Electric field intensity due to electric dipole at equatorial or broad side position
*“Electric field intensity due to electric dipole at equatorial or broad side position”:
*Note: please replace 2a into 2l in fig.
Consider AB is an electric dipole which is consisting by two point charges -q and + q respectively of lenght 2l, thus the electric dipole moment will be ,p = 2l×q . we are to find the electric field intensity due to this dipole in broad side on position at a distance r from midpoint of the dipole at point P for this assuming that the unit positive charge is kept at point P then intensity produced by the + q and -q charges at point p are E1 and E2 respectively. By the vector resolution we find the rectangular component of E1 are E1cosθ and E1sinθ· Similarly, the vector component of E2 are E2cosθ and E2sinθ·
In right ∆AOP By pythgoras
AP² = OP² + OA²
AP² = r² + l²
AP = √r²+l² = (r² +l²)½
But AP = BP
Hence, AP² = BP² = (r+l)
And, cosθ = OA/AP = l/(r²+l²)½
the electric field intensity due to point charge +q is
and the electric field intensity due to point charge -q E2 is same as E1 but the direction is different·
It is clear that E1= E2 · Thus from figure the component E1sinθ and
E2sinθ are equal in magnitude and
*Note: please replace 2a into 2l in fig.
Consider AB is an electric dipole which is consisting by two point charges -q and + q respectively of lenght 2l, thus the electric dipole moment will be ,p = 2l×q . we are to find the electric field intensity due to this dipole in broad side on position at a distance r from midpoint of the dipole at point P for this assuming that the unit positive charge is kept at point P then intensity produced by the + q and -q charges at point p are E1 and E2 respectively. By the vector resolution we find the rectangular component of E1 are E1cosθ and E1sinθ· Similarly, the vector component of E2 are E2cosθ and E2sinθ·
In right ∆AOP By pythgoras
AP² = OP² + OA²
AP² = r² + l²
AP = √r²+l² = (r² +l²)½
But AP = BP
Hence, AP² = BP² = (r+l)
And, cosθ = OA/AP = l/(r²+l²)½
the electric field intensity due to point charge +q is
and the electric field intensity due to point charge -q E2 is same as E1 but the direction is different·
It is clear that E1= E2 · Thus from figure the component E1sinθ and
E2sinθ are equal in magnitude and
opposite in direction· Then they are cancelation each other but the remainig components E1cosθ and
E2cosθ are in same direction· Thus
the resultant intensity is the addition
of E1cosθ and E2cosθ·
But 2ql=p (moment of electric dipole)
The direction of electric field E is 'antiparallel' to the dipole axis.
If r is very large compared to 2l (r» 2l), then , l² ≈ 0
Newton/Coulomb
![Electric field intensity due to charged metallic sphere [solid or hollow]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgCwkEB3m1usMzcmllNbxfYqi6PuvmJNlTnkYxik7Xeow_fBYLxrEfn-xDpGyH9eMYOXFxifmCXDqELg5CmPfl2jFmjCaVQ2sjCNIHg8v0WkKEUa1TvW2884bmteo3yscYTxytRRPDEAEk/s72-c/IMG_20200423_142800.jpg)
No comments: