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Hint: In order to answer this question, to know the potential energy of an electric dipole in an uniform electric field, we will find the potential energy numerically. We will assume each and every important term that is involved in this question.
Complete answer:
as potential energy in the system's electric field (the dipole and the charges responsible for the electric field). Let us consider an electric dipole of the dipole moment $\mathop \to \limits_P $ in a uniform electric field $\mathop \to \limits_E $ with $\mathop \to \limits_P $ making an angle $\theta $ with $\mathop \to \limits_E $.
The torque acting on the dipole
$\mathop \to \limits_T = \mathop \to \limits_P \times \mathop \to \limits_E $
$T = pE\sin \theta $
let $dW$ be the amount of work done in rotating the dipole through an angle $d\theta$. Clearly,
\[dW = pESin\theta d\theta \]
Obviously, $dW$ is the change in potential energy $dU$ of the dipole, i.e.,
\[dW = dW = pESin\theta d\theta \] ……(1)
If the angel changes from \[{90^o}\;to\;\theta \] then the potential energy will become
\[W = \smallint _0^\theta pEsin\theta d\theta \] ……….(2)
$ \Rightarrow W = E( - cos\theta )_0^\theta $
\[ \Rightarrow W = pE\left( {1-Cos\theta } \right)\]
We have considered the value of $\theta $ from \[{90^o}\;to\;\theta \], since the dipole axis is perpendicular to the field so potential energy is equivalent to zero.
\[U({90^o}) = 0\]
Equation (2) becomes-
\[U(\theta )-U({90^\circ }) = \smallint _{{{90}^\circ }}^\theta pEsin\theta d\theta \\
\Rightarrow U(\theta )- 0 = pE( - cos\theta )_{{{90}^\circ }}^\theta \\
\Rightarrow U(\theta ) = - pEcos\theta \\
\therefore U(\theta ) = - p.E\]
Note:When a dipole is positioned in a uniform electric field, the electric field exerts force on the dipole, causing it to rotate clockwise or anticlockwise. The electric field and potential energy of an electric dipole are discussed here. The electric field of an electric dipole is calculated here.
Complete answer:
as potential energy in the system's electric field (the dipole and the charges responsible for the electric field). Let us consider an electric dipole of the dipole moment $\mathop \to \limits_P $ in a uniform electric field $\mathop \to \limits_E $ with $\mathop \to \limits_P $ making an angle $\theta $ with $\mathop \to \limits_E $.
The torque acting on the dipole
$\mathop \to \limits_T = \mathop \to \limits_P \times \mathop \to \limits_E $
$T = pE\sin \theta $
let $dW$ be the amount of work done in rotating the dipole through an angle $d\theta$. Clearly,
\[dW = pESin\theta d\theta \]
Obviously, $dW$ is the change in potential energy $dU$ of the dipole, i.e.,
\[dW = dW = pESin\theta d\theta \] ……(1)
If the angel changes from \[{90^o}\;to\;\theta \] then the potential energy will become
\[W = \smallint _0^\theta pEsin\theta d\theta \] ……….(2)
$ \Rightarrow W = E( - cos\theta )_0^\theta $
\[ \Rightarrow W = pE\left( {1-Cos\theta } \right)\]
We have considered the value of $\theta $ from \[{90^o}\;to\;\theta \], since the dipole axis is perpendicular to the field so potential energy is equivalent to zero.
\[U({90^o}) = 0\]
Equation (2) becomes-
\[U(\theta )-U({90^\circ }) = \smallint _{{{90}^\circ }}^\theta pEsin\theta d\theta \\
\Rightarrow U(\theta )- 0 = pE( - cos\theta )_{{{90}^\circ }}^\theta \\
\Rightarrow U(\theta ) = - pEcos\theta \\
\therefore U(\theta ) = - p.E\]
Note:When a dipole is positioned in a uniform electric field, the electric field exerts force on the dipole, causing it to rotate clockwise or anticlockwise. The electric field and potential energy of an electric dipole are discussed here. The electric field of an electric dipole is calculated here.
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