Question

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

Answer 1

Answer:

A dipole is a pair of opposite charges with equal magnitudes separated by a distance, d. ... Also note that when the angle is 90 degrees, the point P is equidistant to both charges and the electric potential is zero. When θ > 90⁰, the potential is negative because the point P is closer to the negative charge.

Therefore, the electric potential due to an electric dipole at a given point is equal to KPcosθr2−a2cos2θ. ... (i) When the given point is on the axial line of the dipole (i.e. θ=0). Substitute θ=0 in equation (iv). Therefore, V=KPcos0r2−a2cos20.

Answer 2

Answer:

$\huge\star \underline{ \boxed{ \purple{Answer}}}\star$

➝ 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 = $\dfrac{1}{4\pi \:Eo } \dfrac{p}{ {r}^{2} }$

$\huge\mathfrak\pink{Hope \: it \: helps \: you}$

$\huge\mathfrak\pink{friend}$