Question

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ᴅᴇғɪɴᴇ ᴘᴏᴛᴇɴᴛɪᴀʟ ᴅᴜᴇ ᴛᴏ ᴅɪᴘᴏʟᴇ ?
ᴀɴᴅ ᴅᴇʀɪᴠᴇ ᴀɴ ᴇxᴘʀᴇssɪɪɴ ғᴏʀ ɪᴛ ᴀʟsᴏ ?​

Answer 1

$\huge\bf\pink{\mid{\overline{\underline{Answer :- }}}\mid}$

➝ The amount of work done against the electrostatic force to move a unit positive charge from one point to another is known a potential due to electric dipole.

Derivation :-

Potential at +q {V(+q)} = $\frac{1}{4\pi\:Eo} \frac{ + q}{(r - a)}$

Potential at -q {V(-q)} = $\frac{1}{4\pi\:Eo} \frac{ - q}{(r + a)}$

Vp = V+q + V-q ( Total V )

➝ V = $\frac{1}{4\pi\:Eo} \frac{ + q}{(r - a)}$ +

$\frac{1}{4\pi\:Eo} \frac{ - q}{(r + a)}$

➝ V = $\frac{1}{4\pi \: Eo} \times ( \frac{q}{(r - a)} + \frac{( - q)}{(r + a)} )$

➝ V = $\frac{q}{4\pi \:Eo } ( \frac{1}{(r - a)} - \frac{1}{(r + a)} )$

➝ V = $\frac{q}{4\pi \: Eo} ( \frac{(r + a) - (r - a)}{(r - a)(r + a)} )$

➝ V = $\frac{q}{4\pi \:Eo } ( \frac{r + a - r + a}{(r - a)(r + a)} )$

➝ V = $\frac{q}{4\pi \:Eo } \frac{2a}{( {r}^{2} - {a}^{2} )}$

We know that,

| p = 2aq |

So,

➝ V = $\frac{1}{4\pi \:Eo } \frac{p}{( {r}^{2} - {a}^{2} )}$

r² - a² ≈ r²

Where, short dipole

Where, short dipole a² ➝ 0 , w.r.t r , [ r >> a ]

Hence, V = $\dfrac{1}{4\pi \:Eo } \dfrac{p}{ {r}^{2} }$

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Answer 2

may this answer helps you ....=_=