Question

Two charged particles each of mass $9.8g$ and charges $+20\mu C$ and $-20\mu C$ are attached to the two ends of a massless and rigid uniform non-conducting rod of length $50cm$. This arrangement is held in a uniform electric field of $12.1N{C}^{-1}$, such that the rod makes a very small angle with the field direction. If the rod is set free, the minimum time needed for the rod to become parallel to the direction of the electric field is ............... seconds.

• A. 5
• B. 8
• C. 12
• D. 17

Answer 1

Answer: (A)

This arrangement of electric dipole in electric field will undergo SHM, when the rod is set free. Time period of this SHM is given by
$T=2\pi \sqrt{\frac{I}{PE}}$
where, $I=$ moment of inertia of dipole,
$p=$ electric dipole moment and
$E=$ electric field.
$I=2m{r}^2$
mass, $m=9.8\times 10^{-3}kg,r=\frac{50}{2}\times 10^{-2}$
$I=2\times 9.8\times 10^{-3}\times {\left(\frac{50}{2}\times 10^{-2}\right)}^2$
$I=1225\times 10^{-4}k-g{m}^2$
$p=q(2l)=20\times 10^{-6}\times 50\times 10^{-2}=10^{-5}C-m$
$E=121N/C$
$T=2\pi \sqrt{\frac{1225\times 10^{-4}}{10^{-5}\times 121}}=2\times 3.14\times 3.181$
$T=19.98\cong 20s$
But time taken by the rod to become parallel will be $T/4$, so required time is $5s$.