what is the work done by an electric field in rotating a dipole from its unstable equilibrium to stable equilibrium? does the energy of dipole increase or decrease in this process why?

Potential energy of dipole is the energy possessed by the dipole by virtue of its particular position in the electric field.
Total work done in rotating the dipole from orientation θ1 and θ2 is stored in the form of potential energy and is given by :-
W = -pE (cosθ2 - cosθ1)
Where θ1 is the angle at which dipole is in unstable equilibrium = 1800
 
and θ2 is the angle at which dipole is in stable equilibrium = 00

Hence, work done , W = -2pE and hence the energy of the dipole decreases in this process.

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Work done = 2pE.

Potential energy of the dipole decreases.

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WHY? doesnt the energy remain same?

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Energy DOES NOT remain the same.

Potential energy is given U = -p.E = -pEcos(theeta).

In stable eq, theeta = 0

In unstable eq, theeta = pi.

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