Question

Derive the electric field at an axial point.

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Answer 1

`\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}`

before looking at what the question says ,

let's have some general information related to the topic of the question !

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`\implies \:` Electric field refers to the force generated by a nearby charge in the surroundings.

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`\implies \:` Mathematically ,

Electric field =
`(Kq)/(r^(2)) \\`

where ,

K = coulomb's constant

q = magnitude of charge

r = distance between the opposite terminals

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`\implies \:` Dipole moment , P = 2qd

where ,

q = magnitude of charges

d = distance between the charges

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`\implies \:` positive charge tends to throw the field away from it while negative charge tends to pull the field towards it !

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seems like enough of information , so now let's look at the derivation !

refer the attachments in order to see the derivation !

Final answer -

`\bold\pink{E = \frac{2kp}{(x {}^(2) - d {}^(2) ) {}^(2) } } \\`

while , for a short dipole

`\bold\pink{E = \frac{2pk}{x {}^(3) }} \\`

hope helpful :D