Advertisements
Advertisements
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.
Advertisements
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 )
APPEARS IN
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)`
Define dipole moment of an electric dipole. Is it a scalar or a vector?
Two particles A and B, of opposite charges 2.0 × 10−6 C and −2.0 × 10−6 C, are placed at a separation of 1.0 cm. Calculate the electric field at a point on the perpendicular bisector of the dipole and 1.0 m away from the centre.
Two particles, carrying charges −q and +q and and of mass m each, are fixed at the ends of a light rod of length a to form a dipole. The rod is clamped at an end and is placed in a uniform electric field E with the axis of the dipole along the electric field. The rod is slightly tilted and then released. Neglecting gravity, find the time period of small oscillations.
Answer the following question.
Derive an expression for the electric field at any point on the equatorial line of an electric dipole.
An electric dipole consists of two opposite charges each 0.05 µC separated by 30 mm. The dipole is placed in an unifom1 external electric field of 106 NC-1. The maximum torque exerted by the field on the dipole is ______
When an electric dipole p is placed in a uniform electric field E then at what angle the value of torque will be maximum?
A conic surface is placed in a uniform electric field E as shown in the figure such that the field is perpendicular to the surface on the side AB. The base of the cone is of radius R, and the height of the cone is h. The angle of the cone is θ.

Find the magnitude of the flux that enters the cone's curved surface from the left side. Do not count the outgoing flux (θ < 45°)
The ratio of the weight of a man in a stationary lift and in a lift accelerating downwards with a uniform acceleration α is 3 : 2. The acceleration of the lift is:
A square surface of side l (m) in the plane of the paper. A uniform electric field E(V/m) also in the plane of the paper is limited only to the lower half of the square surface, the electric flux (in SI units) associated with the surface is ______.

