Question
When electric dipole is placed in uniform electric field, its two charges experience equal and opposite forces, which cancel each other and hence net force on electric dipole in uniform electric field is zero. However, these forces are not collinear, so they give rise to some torque on the dipole. Since net force on electric dipole in uniform electric field is zero, so no work is done in moving the electric dipole in uniform electric field. However, some work is done in rotating the dipole against the torque acting on it.
- The dipole moment of a dipole in a uniform external field $\vec{\text{E}}$ is $\vec{\text{P}}.$ Then the torque $'i $ acting on the dipole is:
- $\text{pE}\sin\theta,-\text{pE}\cos\theta$
- $\text{pE}\sin\theta,-2\text{pE}\cos\theta$
- $\text{pE}\sin\theta,2\text{pE}\cos\theta$
- $\text{pE}\cos\theta,-2\text{pE}\sin\theta$
- An electric dipole of moment pis placed in an electric field of intensity $E.$ The dipole acquires a position such that the axis of the dipole makes an angle $\theta$ with the direction of the field. Assuming that the potential energy of the dipole to be zero when $\theta=90^\circ$ the torque and the potential energy of the dipole will respectively be:
- $\text{F}=0, \tau=0$
- $\text{F}\not=0, \tau\not=0$
- $\text{F}=0, \tau\not=0$
- $\text{F}\not=0, \tau=0$
- When an electric dipole is held at an angle in a uniform electric field, the net force $F$ and torque t on the dipole are:
- $0^\circ$
- $90^\circ$
- $180^\circ$
- Both $(a)$ and $(c)$
- Torque on a dipole in uniform electric field is minimum when $\theta$ is equal to:
- $0.2 \times 10^{-3}Nm$
- $1 \times 10^{-3}Nm$
- $2 \times 10^{-3}Nm$
- $4 \times 10^{-3}Nm$
- An electric dipole consists of two opposite charges, each of magnitude $1.0\mu\text{C}$ separated by a distance of $2.0\ cm$.
- The dipole is placed in an external field of $10^5N C^{-1}$.
- The maximum torque on the dipole is:
- $\vec{\tau}=\vec{\text{P}}\times\vec{\text{E}}$
- $\vec{\tau}=\vec{\text{P}}\cdot\vec{\text{E}}$
- $\vec{\tau}=2(\vec{\text{P}}+\vec{\text{E}})$
- $\vec{\tau}=(\vec{\text{P}}+\vec{\text{E}})$
