Physics, asked by mohith73821, 4 months ago

electric field due to dipole moment P on the on its axial line at a distance r from the centre​

Answers

Answered by gumedha0709
0

Explanation:

Electric field due to an electric dipole at a point on its axial line: AB is an electric dipole of two point charges −q and +q separated by small distance 2d. P is a point along the axial line of the dipole at a distance r from the midpoint O of the electric dipole.

The electric fiedl at the point P due to +q placed at B is,

E  

1

​  

=  

4πε  

0

​  

 

1

​  

 

(r−d)  

2

 

q

​  

 (along BP)

The electric field at the point P due to −q placed at A is,

E  

2

​  

=  

4πε  

0

​  

 

1

​  

 

(r+d)  

2

 

q

​  

 (along PA)

Therefore, the magnitude of resultant electric field (E) acts in the direction of the vector with a greater, magnitude. The resultant electric field at P is

E=E  

1

​  

+(−E  

2

​  

)

E=[  

4πε  

0

​  

 

1

​  

 

(r−d)  

2

 

q

​  

−  

4πε  

0

​  

 

1

​  

 

(r+d)  

2

 

q

​  

] along BP

E=  

4πε  

0

​  

 

q

​  

[  

(r−d)  

2

 

1

​  

−  

(r+d)  

2

 

1

​  

] along BP

E=  

4πε  

0

​  

 

q

​  

[  

(r  

2

−d  

2

)  

2

 

4rd

​  

] along BP

If the point P is far away from the dipole, then d≪r

∴E=  

4πε  

0

​  

 

q

​  

−  

r  

4

 

4rd

​  

=  

4πε  

0

​  

 

q

​  

 

r  

3

 

4d

​  

 

E=  

4πε  

0

​  

 

1

​  

 

r  

3

 

2p

​  

 along BP

            [∵ Electric dipole moment p=q×2d]

E acts in the direction of dipole moment.

Answered by soniatiwari214
0

Answer:

The electric field due to dipole moment P on its axial line at a distance r from the centre​ is E = 1/4πε₀×2p/r³

Explanation:

  • At a point on its axial line, an electric dipole produces an electric field: Two point charges, q and +q, separated by a short distance 2d form the electric dipole known as AB.
  • P is a point on the dipole's axial line located r units away from the electric dipole's midpoint O.
  • The electric field at point P caused by the placement of +q at B is,
  • E₁ = 1/4πε₀×q/(r-d)² along BP
  • When -q is positioned at point A, the electric field at point P results in E₂ = 1/4πε₀×q/(r+d)²  (along PA)
  • As a result, the resultant electric field (E) has a larger magnitude and acts in the direction of the vector. The electric field at P as a result is, E = E₁ + (-E₂)
  • E = 1/4πε₀×q/(r-d)² - 1/4πε₀×q/(r+d)²  along BP
  • E = q/4πε₀ [1/(r-d)² - 1/(r+d)²] along BP
  • E = q/4πε₀ [4rd/ (r²-d²)²] along BP
  • Point P will be far from the dipole if d<<r.
  • Therefore, E = q/4πε₀ - 4rd/r⁴ = q/4πε₀ - 4d/r³
  • Since electric dipole moment p = q×2d
  • Therefore, the equation of electric field becomes-
  • E = 1/4πε₀×2p/r³
  • Here, E acts towards the dipole moment direction.

Thus, the electric field due to dipole moment P on its axial line at a distance r from the centre​ is E = 1/4πε₀×2p/r³

#SPJ3

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