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Calculate the electric field due to a dipole on its axial line and equatorial plane.

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#### Solution

**Case (i):**

Electric field due to an electric dipole at points on the axial line.

- Consider an electric dipole placed on the x-axis A point C is located at a distance of r from the midpoint of the dipole along the axial line.
**Electric field of the dipole along the axial line** - The electric field at a point C due to +q is

`vec"E"_+ = 1/(4 pi ε_0) "q"/("r - a")^2` (along BC) - Since the electric dipole moment vector p is from -q to +q and is directed along BC, the above equation is rewritten as

`vec"E"_+ = 1/(4 pi ε_0) "q"/("r - a")^2 hat"P"` ....(1) - 4. The electric field at a point C due to -q is

`vec"E"_- = -1/(4 pi ε_0) "q"/("r + a")^2 hat"P"` ....(2) - Since +q is located closer to the point C than -q, `vec"E"_+` is stronger than `vec"E"_-`.

Therefore, the length of the `vec"E"_+` vector is drawn larger than that of `vec"E"_-` vector.

The total electric field at point C is calculated using the superposition principle of the electric field.

`vec"E"_"tot" = vec"E"_+ + vec"E"_-`

`= 1/(4pi ε_0) "q"/("r - a")^2 hat"p" - 1/(4pi ε_0) "q"/("r + a")^2 hat"p"`

`vec"E"_"tot" = "q"/(4pi ε_0) (1/("r - a")^2 - 1/("r + a")^2)hat"p"` .....(3)

`vec"E"_"tot" = "q"/(4pi ε_0) ("4ra"/("r"^2 - "a"^2)^2) hat"p"` .....(4)

r >> a

hence (r^{2}- a^{2})^{2}≈ r^{4}Substitute in eqn (4)

`vec"E"_"tot" = 1/(4pi ε_0)("4aq"/"r"^3)hat "p"`

`2"aq"hat"p" = vec"p"`

`vec"E"_"tot" = 1/(4pi ε_0) (2vac"p")/"r"^3` - The direction of is shown in Figure
**Total electric field of the dipole on the axial line**

**Case (ii)**

- Electric field due to an electric dipole at a point on the equatorial plane.
- Consider a point C at a distance r from the midpoint 0 of the dipole on the equatorial plane.
- Since point C is equidistant from +q and -q and are the same.
- The direction of `vec"E"_+` is along BC and the direction of `vec"E"_-` is along CA.

`vec"E"_+` and `vec"E"_-` are resolved into two components; one component parallel to the dipole axis and the other perpendicular to it. - The perpendicular components `|vec"E"|` sinθ and `|vec"E"_-|` sinθ are oppositely directed and cancel each other.

`"E"_"tot" = - |"E"_+| cos theta hat "p" - |vec"E"_-| cos theta hat "p"` .....(6) - The magnitudes `vec"E"_+ and vec"E"_-` are the same and are given by
- By substituting equation (7) into equation (6) we get

`vec"E"_"tot" = - 1/(4piε_0) (2"q" cos theta)/("r"^2 + "a"^2) hat"p"`

since cos θ = `"a"/sqrt("r"^2 + "a"^2) hat`

=`- 1/(4piε_0) (2"qa")/("r"^2 +"a"^2)^(3/2) hat"p"`

since `vec"p" = "2qa"hat"p"` ....(8)

`vec"E"_"tot" = - 1/(4piε_0) vec"p"/("r"^2 +"a"^2)^(3/2)`

`|"E"_+| = |vec"E"_-| = 1/(4piε_0) "q"/("r"^2 + "a"^2)` .....(7) - At very large distances (r»a), the equation (8) becomes

`vec"E"_"tot" = - 1/(4piε_0) vec"p"/"r"^3` (r >> a)

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