Question

Electric field at a point on the axis of a dipole derivation

Answer 1

By using the formula for electric field due to point charge,

Electric field due to +q =
+1
4πε0

q
(r - l)2

The distance between (P and +q) = (r-l)

Electric field due to -q =
-1
4πε0

q
(r + l)2

The distance between (P and -q) = (r + l)

(Electric field due to +q will be positive and electric field due to- q will be negative)

Since electric field is a vector quantity so, the net electric field will be the vector addition of the two.

So, the net electric field E = E1 + E2

E =
1
4πε0

q
(r - l)2
-
1
4πε0

q
(r + l)2

E =
q
4πε0
[
1
(r - l)2
-
1
(r + l)2
]

On solving the equation we get −

E =
q
4πε0
[
(r + l)2 - (r - l)2
(r - l)2(r + l)2
]

E =
q
4πε0

4rl
(r2 - l2)2
......(1)

We know that the dipole moment or effectiveness of dipole (P) is given by −

P = 2ql

Therefore, putting this value in eq(1), we get

E =
1
4πε0

2Pr
(r2 - l2)2
......(2)

Certain assumptions are made based on this equation −

Since, the dipole is very small so ‘l’ is also very small as compared to the distance ‘r’.

So, on neglecting ‘r’ with respect to ‘l’ we get −

E =
1
4πε0

2Pr
r4
(from eq(2))

⇒ E =
1
4πε0

2P
r2

E =
1
4πε0

2P
r2