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प्रश्न
Define electric dipole moment. Is it a scalar or a vector? Derive the expression for the electric field of a dipole at a point on the equatorial plane of the dipole.
उत्तर
Electric dipole moment: The strength of an electric dipole is measured by the quantity electric dipole moment. Its magnitude is equal to the product of the magnitude of either charge and the distance between the two charges.
Electric dipole moment, p = q × d
It is a vector quantity.
In vector form it is written as`vecp = qxxvecd`, where the direction of `vecd`is from negative charge to positive charge.
Electric Field of dipole at points on the equatorial plane:
The magnitudes of the electric field due to the two charges +q and −q are given by,
`E_(+q) = q/(4piε_0) 1/(r^2+a^2) ...... (1)`
`E_(-q) = q/(4piε_0) 1/(r^2+a^2) ...... (2)`
`therefore E_(+q) =E_(-q)`
The directions of E+q and E−q are as shown in the figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up.
∴ Total electric field
`E = -(E_(+q) + E_(-q)) cos theta hatp`
[Negative sign shows that field is opposite to `hatp` ]
`E = -(2qa)/(4piε_0 r^3)hatP ........ (3)`
At large distances (r >> a), this reduces to
`E = -(2qa)/(4piε_0 r^3)hatP ....... (4)`
`because vecp = q xx 2veca hatp`
`because E = (-vecp)/(4piε_0 r^3)` ( r>>a )
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संबंधित प्रश्न
An electric dipole of dipole moment`vecp` consists of point charges +q and −q separated by a distance 2a apart. Deduce the expression for the electric field `vecE` due to the dipole at a distance x from the centre of the dipole on its axial line in terms of the dipole moment `vecp`. Hence show that in the limit x>> a, `vecE->2vecp"/"(4piepsilon_0x^3)`
A short electric dipole (which consists of two point charges, +q and -q) is placed at the centre 0 and inside a large cube (ABCDEFGH) of length L, as shown in Figure 1. The electric flux, emanating through the cube is:
a) `q"/"4piin_9L`
b) zero
c) `q"/"2piin_0L`
d) `q"/"3piin_0L`
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