English, asked by payal1goswami2, 9 months ago

derive an expression for electric field at a point on the axis of dipole​

Answers

Answered by adityasuyal180164
0

Explanation:

Derive an expression for the electric field intensity at a point on the axis of an electric dipole. ... A system has two charges qA = 2.5 x 10–7 C and qB = –2.5 x 10–7 C located at points A: (0, 0, –15 cm) and B: (0, 0, + 15 cm), respectively.

Answered by jainamshah1595
1

Answer:

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.

Explanation:

hope it helps....

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