Question

The linear charge density of a thin rod of length $L$ is given by $\lambda =\left({\lambda }_{0}x\right)\text{Coulomb/m},[0\le x<\frac{L}{2}]$ and $\lambda =\left({\lambda }_0{x}^2\right)\text{Coulomb/m},[\frac{L}{2}\le x\le L]$. Find the total charge on rod. $(x$ is measure from one end of the rod$)$.

• A. $\frac{{\lambda }_0{L}^2}{24}(3+7L)$
• B. $\frac{{\lambda }_0{L}^3}{8}(1+5L)$
• C. $\frac{{\lambda }_0{L}^2}{4}(1+7L)$
• D. ${\lambda }_0{L}^2(1+7L)$

Answer 1

The correct option is A $\frac{{\lambda }_0{L}^2}{24}(3+7L)$

Total charge on $\text{AB}$ can be calculated by,

$q_{AB}=\underset{x=0}{\overset{x=\frac{L}{2}}{\int }}\lambda dx$

From the data given in the question,

$q_{AB}=\underset{x=0}{\overset{x=\frac{L}{2}}{\int }}{\lambda }_{0}xdx$

$\Rightarrow q_{AB}={\lambda }_0{\left[\frac{x^2}{2}\right]}_{x=0}^{x=\frac{L}{2}}$

$\therefore q_{AB}=\frac{{\lambda }_0{L}^2}{8}$

Similarly , total charge on $\text{BC}$ can be calculated by,

$q_{BC}=\underset{x=\frac{L}{2}}{\overset{x=L}{\int }}\lambda dx$

$\Rightarrow q_{BC}=\underset{x=\frac{L}{2}}{\overset{x=L}{\int }}{\lambda }_0{x}^{2}dx$

$\Rightarrow q_{BC}=\frac{{\lambda }_0}{3}[x^3{]}_{x=\frac{L}{2}}^{x=L}$

$\Rightarrow q_{BC}=\frac{{\lambda }_0}{3}[L^3-\frac{L^3}{8}]$

$\Rightarrow q_{BC}=\frac{7{\lambda }_0{L}^3}{24}$

Total charge on $\text{AC}$,

$q=q_{AB}+q_{BC}$

$\Rightarrow q=\frac{{\lambda }_0{L}^2}{8}+\frac{7{\lambda }_0{L}^3}{24}$

$\Rightarrow q=\frac{{\lambda }_0{L}^2}{24}(3+7L)$

Hence, option (a) is the correct answer.