Solution:

Ans. Here, $m=10^{-3} \mathrm{~kg}, q=5 \times 10^{-6} \mathrm{C}, E=2 \times 10^{5} \mathrm{~N} / \mathrm{C}$, $u=20 \mathrm{~m} / \mathrm{s}, v=0$

As the particle enters opposite to the field, so it will retard.

Acceleration

$$ \begin{aligned} a & =-\frac{q \mathrm{E}}{m} \ & =-\frac{5 \times 10^{-6} \times 2 \times 10^{5}}{10^{-3}} \end{aligned} $$

$$ =-10^{3} \mathrm{~m} / \mathrm{s}^{2} $$

Using,

$$ v^{2}=u^{2}-2 a s $$

$0=(20)^{2}-2 \times 1000 \times s$

$\Rightarrow \quad s=\frac{400}{2000}=\frac{1}{5}=0.2 \mathrm{~m}$

1

[CBSE Marking Scheme, 2012]

Unstable equilibrium $\theta=180^{\circ}$

[CBSE Marking Scheme, 2017]

(AI) Q. 2. (i) Obtain the expression for the torque $\vec{\tau}$ experienced by an electric dipole of dipole moment $\vec{p}$ in a uniform electric field, $\vec{E}$.

(ii) What will happen if the field were not uniform?

R [Delhi III 2017]

Ans. (i) Obtaining expression for torque $\vec{\tau}$ experienced by electric dipole in uniform electric field 2

(ii) Effect of non-uniform electric field

(i) Force on $+q, \vec{F}=q \vec{E}$

Force on $-q, \vec{F}=-q \vec{E}$

Magnitude of torque $\quad \vec{\tau}=q E \times 2 a \sin \theta$

$$ =2 q a E \sin \theta $$

$$ \vec{\tau}=\vec{p} \times \vec{E} $$

(ii) If the electric field is non-uniform, the dipole experiences a translatory force as well as a torque.

[CBSE Marking Scheme, 2017]

Detailed Answer :

(i)

Consider electric dipole kept in an uniform electric field at an angle $\theta$ where a dipole experience a torque, so, torque generated by parallel forces $q E$ will act as couple as

$$ \begin{aligned} |\vec{\tau}| & =q E 2 l \sin \theta \ & =p E \sin \theta \quad[\text { as } p=2 q l] 1 / 2 \ |\vec{\tau}| & =|\vec{p} \times \vec{E}| \end{aligned} $$

(ii) When the field is non-uniform, force acting on both ends will not be equal, hence they result in mixture of couple and net force. With this, dipole experiences rotational as well as linear force. $\mathbf{1}$