An electric dipole of dipole moment p is lying along uniform electric field E. Calculate the work done in rotating the dipole by 90 degrees.

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Ques: An electric dipole of dipole moment p is lying along uniform electric field E. Calculate the work done in rotating the dipole by 90 degrees.

Ans: The work done in rotating the dipole by 90 degrees when dipole moment is lying along uniform electric field is “pE”.

Explanation: The potential energy of an electric dipole in a uniform electric field is given by an expression: U = -pEcosθ.

where p is the dipole moment,
E is the electric field strength,
θ is the angle between the dipole moment and the electric field.

Initially, the dipole moment is aligned with the electric field, so θ = 0, and the potential energy is:

U1 = -pEcos0 = -pE

When the dipole is rotated by 90 degrees, the angle between the dipole moment and the electric field becomes θ = 90 degrees, and the potential energy becomes:

U2 = -pEcos90 = 0

The change in potential energy is then equal to the work done

Work Done = ΔU = U2 - U1 = 0 - (-pE) = pE

electric dipole of dipole moment p is lying along uniform electric field E

Therefore, the potential energy of the dipole in rotating it by 90 degrees is pE.

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An electric dipole of dipole moment p is lying along uniform electric field E. Calculate the work done in rotating the dipole by 90 degrees.

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2.
Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

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5.
A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.

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An electric dipole of dipole moment p is lying along uniform electric field E. Calculate the work done in rotating the dipole by 90 degrees.

CBSE CLASS XII Related Questions

2.Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

4.This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?

5.A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

2.
Derive an expression for the capacitance of a parallel plate capacitor of plate area A and plate separation d with air present between the plates.

4.
This ‘average velocity’ is found be few mm/s for currents in range of a few amperes. How then is current established almost the instant a circuit is closed ?

5.
A student sets up the circuit as shown in the figure to find the value of unknown resistance X and records a set of readings of the voltmeter and the ammeter by using the rheostat.