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प्रश्न
Derive an expression for the electric potential due to an electric dipole.
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उत्तर
- Consider an electric dipole. Let origin be at the centre of the dipole as shown in the figure below.
Electric potential due to an electric Dipole - Let C be any point near the electric dipole at a distance r from the centre O inclined at an angle θ with the axis of the dipole.
Let r1 and r2 be the distances of point C from charges +q and –q, respectively. - Potential at C due to charge +q at A is,
V1 = `(+"q")/(4piε_0"r"_1)`
Potential at C due to charge –q at B is,
V2 = `(-"q")/(4piε_0"r"_2)` - The potential at C due to the dipole is,
`"V"_"C" = "V"_1 + "V"_2 = "q"/(4piε_0)[1/"r"_1 - 1/"r"_2]` ….(1) - By geometry,
`"r"_1^2 = "r"^2 + "l"^2 - 2"rl" "cos"θ`
`"r"_2^2 = "r"^2 + "l"^2 + 2"rl" "cos"θ`
`"r"_1^2 = "r"^2(1 + "l"^2/"r"^2 - 2"l"/"r" cosθ)`
`"r"_2^2 = "r"^2(1 + "l"^2/"r"^2 + 2"l"/"r" cosθ)`
For a short dipole, 2l << r and
If r >> l; `"l"/"r"` is small
∴ `"l"^2/"r"^2` can be neglected
∴ `"r"_1^2 = "r"^2(1 - 2"l"/"r" cosθ)`
`"r"_2^2 = "r"^2(1 + (2"l")/"r" cosθ)`
∴ `"r"_1 = "r"(1 - (2l)/"r" cosθ)^{1/2}`
∴ `"r"_2 = "r"(1 + (2l)/"r" cosθ)^{1/2}`
∴ `1/"r"_1 = 1/"r"(1 - (2l)/"r"cosθ)^{(-1)/2}` and
`1/"r"_2 = 1/"r"(1 + (2l)/"r"cosθ)^{(-1)/2}` ….(2) - Using equations (1) and (2),
VC = V1 + V2
= `"q"/(4piepsilon_0)[1/"r"(1 - (2lcosθ)/"r")^{(-1)/2} - 1/"r"(1 + (2lcosθ)/"r")^{(-1)/2}]` - Using binomial expansion,
(1 + x)n = 1 + nx, x << l and retaining terms up to the first order of `l/"r"` only, we get
VC = `"q"/(4piepsilon_0) 1/"r" [(1 + l/"r" cosθ) - (1 - l/"r" cosθ)]`
= `"q"/(4piepsilon_0"r") [1 + l/"r" cosθ - 1 + l/"r" cosθ]`
= `"q"/(4piepsilon_0"r")["2l"/"r" cosθ]`
∴ `"V"_"C" = 1/(4piepsilon_0) ("p"cosθ)/"r"^2` ......(∵ p = q × 2l) - Electric potential at C, can also be expressed as,
`"V"_"C" = 1/(4piepsilon_0) (vec"p".vec"r")/("r"^3)`
`"V"_"C" = 1/(4piepsilon_0)(vec"p".hat"r")/("r"^2)`, `(hat"r" = vec"r"/"r")` .........`(hat"r" = vec"r"/"r")`
where `hat"r"` is a unit vector along the position vector `vec("OC") = hat"r"`
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