Question

A thin insulating rod of mass $m$ and length $L$ is hinged at its upper end $\left(O\right)$ so that it can freely rotate in vertical plane. The linear charge density on the rod varies with distance $\left(y\right)$ measured from upper end as
$$\lambda =\{\begin{array}{cc}a{y}^2& \text{for}0\le y\le \frac{L}{2}\\ -b{y}^n& \text{for}\frac{L}{2}\le y\le L\end{array}$$ Where $a$ and $b$ both are positive constants. when a horizontal electric field $E$ is switched on the rod is found to remain stationary. Find the value of $\frac{a}{b}$. (Give integer value)

Answer 1

From the given figure, it is evident that, from $y=0$ to $y=\frac{L}{2}$ the rod has positive charge. Force on it $\left(F_1\right)$ is towards right due to electric field.

The lower half has negative charge and electric force $\left(F_2\right)$ on it is towards left.

The torque due to $F_1$ and $F_2$ must balance. $${\tau }_{F1}={\tau }_{F2}$$ Let is consider an infinitesimal element of length $dy$ at a distance $y$ from hinge in the upper half of the rod.

Torque due to applied electric field about hinge is ${\tau }_{F1}=\underset{0}{\overset{\frac{L}{2}}{\int }}E\left(a{y}^{2}dy\right)y$

$\Rightarrow {\tau }_{F1}=\frac{a{L}^4}{2^6}......\left(1\right)$

Let is consider an infinitesimal element of length $dy$ at a distance $y$ from hinge in the lower half of the rod.


Torque due to applied electric field about hinge is ${\tau }_{F2}=\underset{0}{\overset{\frac{L}{2}}{\int }}E\left(b{y}^{n}dy\right)y$

$\Rightarrow {\tau }_{F2}=\frac{b}{n+2}(1-\frac{1}{2^{n+2}})......\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$ by comparing the power of $L$ we get , $n=2$

Substituting this we get,

$\frac{a}{2^6}=\frac{b}{4}\cdot (1-\frac{1}{2^4})\Rightarrow \frac{a}{b}=15$

Accepted answer : $15$