The potential energy of the dipole in the equilibrium position is :

$+ 3 \times 10^{- 23} J$
$- 3 \times 10^{- 23} J$
$- 6 \times 10^{- 23} J$
$- 3 \times 10^{- 36} J$

Question

The potential energy of the dipole in the equilibrium position is -3 times 10-23-J-3 times 10-23-J-6 times 10-23-J-3 times 10-36-J

Toppr Admin · Accepted Answer

As dipole is in equilibrium P-E- should be minimum-x2234-x3B8- angle between dipole and electric field -0oNow- potential energyU-x2212-P-E-x2212-7-68-xD7-10-x2212-29-xD7-4-xD7-105-x2212-3-xD7-10-x2212-23-xA0-J

The potential energy of the dipole in the equilibrium position is :+3×10−23J−3×10−23J−6×10−23J−3×10−36J

As dipole is in equilibrium P.E. should be minimum∴θ= angle between dipole and electric field =0oNow, potential energyU=−P.E=−7.68×10−29×4×105=−3×10−23 J

The potential energy of the dipole in the equilibrium position is :

$+ 3 \times 10^{- 23} J$
$- 3 \times 10^{- 23} J$
$- 6 \times 10^{- 23} J$
$- 3 \times 10^{- 36} J$

As dipole is in equilibrium P.E. should be minimum
$∴ θ =$ angle between dipole and electric field $= 0^{o}$
Now, potential energy
$U = - P . E$
$= - 7.68 \times 10^{- 29} \times 4 \times 10^{5}$
$= - 3 \times 10^{- 23} J$