Advertisements
Advertisements
Question
Drive the expression for electric field at a point on the equatorial line of an electric dipole.
Advertisements
Solution
Electric Field for 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/(4piepsilon_0)1/(r^2+a^2)` .....(i)
`E_(-q)=q/(4piepsilon_0)1/(r^2-a^2)` .....(ii)
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)/(4piepsilon_0(r^2+a^2)^(3/2))hatp` .....(iii)
At large distances (r >> a), this reduces to
`E=-(2qa)/(4piepsilon_0(r^3))hatp` .....(iv)
`because vecp=qxxvec(2a)hatp`
`therefore E=(-vecp)/(4piepsilon_0(r^3))` (r >> a)
APPEARS IN
RELATED QUESTIONS
Derive the expression for the electric potential due to an electric dipole at a point on its axial line.
Write the expression for the torque \[\vec{\tau}\] acting on a dipole of dipole moment \[\vec{p}\] placed in an electric field \[\vec{E}\].
Find the resultant electric field due to an electric dipole of dipole moment, 2aq, (2a being the separation between the charges ±± q) at a point distant 'x' on its equator.
In which orientation, a dipole placed in a uniform electric field is in (i) stable, (ii) unstable equilibrium?
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 axis of the dipole 1.0 cm away from the centre.
Two charges + 3.2 x 10-19 C and --3.2 x 10-19 C placed at 2.4 Å apart to form an electric dipole. lt is placed in a uniform electric field of intensity 4 x 105 volt/m. The electric dipole moment is ______.
Dimensions of mass in electric field and in electric dipole moment are respectively.
The unit of electric dipole moment is ______.
Polar molecules are the molecules ______.
