Question

Find the electric field intensity due to an electric dipole on its centre

Answer 1

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

Electric field due to +q = +14πε0 q(r - l)2

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

Electric field due to -q = -14πε0 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 = 14πε0 q(r - l)2 - 14πε0 q(r + l)2

E = q4πε0 [1(r - l)2 - 1(r + l)2]

On solving the equation we get −

E = q4πε0[(r + l)2 - (r - l)2(r - l)2(r + l)2]

E = q4πε0 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 = 14πε0 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 = 14πε0 2Prr4 (from eq(2))

⇒ E = 14πε0 2Pr2

E = 14πε0 2Pr2

hope it will help u