Question

An electric dipole consists of small charged objects $A$ and $B$ of charges $-q$ and $+q$ and masses $m$ and $4m$ respectively. They are connected by a light non -conducting rod of length $L$. This system is hinged at $A$ so that it can rotate in vertical plane. A uniform electric field of intensity $E$ is applied vertically downward. The rod is released from horizontal position as shown in figure. The angular velocity of the rod when the rod becomes vertical is.

• A. $\sqrt{\frac{4mg+qE}{2ml}}$
• B. $\sqrt{\frac{6mg+qE}{2ml}}$
• C. $\sqrt{\frac{8mg+qE}{6ml}}$
• D. $\sqrt{\frac{2mg+qE}{2ml}}$

Answer 1

The correct option is A $\sqrt{\frac{4mg+qE}{2ml}}$

The charge of $A=-q$

the charge of $B=+q$

mass of $H_q=4m$

length $=L$

intensity $=E$

angular velocity of the rod. when the rod becomes vertical$=$

$V\times \left(2ml\right)=4mg+qE$

where ($F=4mg$ (force Newton's law)

charge $=q^1$

Now, the force in charge $=qE$ (for coulomb's law)

Hence,

momentum $=V^2\times 2ml$ equating this and we get,

$V^2\times 2ml=4mg+qE$

or, $V^2=\frac{4mg+qE}{2ml}$

or, $V=\sqrt{\frac{4mg+qE}{2ml}}$