- #1
rajeshmarndi
- 319
- 0
When the dipole makes an angle with an electric field, its potential energy is given as the work done in bringing the dipole from the position, when it is aligned with the electric field direction, to a desired position, i.e
W = pE [-cosθf -(-cosθi)] ...eq(1)
p = moment of the dipole
θi = initial position which is when the dipole is aligned with electric field i.e θi = 0
θf = final position of the dipole.
i.e eq(1) becomes,
W= pE(1-cosθf) ...eq(2)
But for convenience, θi is taken when the dipole is perpendicular to electric field, i.e θi= 90°
. This is where I didn't understand. In book, the explanation is just that , it is convenient to take θi=90°, as cos 90°=0, but why and how there is no explanation.
i.e eq(1) now becomes,
W = -pE cosθf ...eq(3)
eq(2) and eq(3) gives different values, so how could they both give the same potential energy of the dipole in an electric field.
W = pE [-cosθf -(-cosθi)] ...eq(1)
p = moment of the dipole
θi = initial position which is when the dipole is aligned with electric field i.e θi = 0
θf = final position of the dipole.
i.e eq(1) becomes,
W= pE(1-cosθf) ...eq(2)
But for convenience, θi is taken when the dipole is perpendicular to electric field, i.e θi= 90°
. This is where I didn't understand. In book, the explanation is just that , it is convenient to take θi=90°, as cos 90°=0, but why and how there is no explanation.
i.e eq(1) now becomes,
W = -pE cosθf ...eq(3)
eq(2) and eq(3) gives different values, so how could they both give the same potential energy of the dipole in an electric field.