Question

If the linear charge density of a rod of length $3\text{m}$ varies as $\lambda =(2+x)\frac{\text{C}}{\text{m}}$, where $x$ is the distance from one end of the rod, then find the total charge on the rod.

• A. $21\text{C}$
• B. $5.5\text{C}$
• C. $10.5\text{C}$
• D. $5\text{C}$

Answer 1

The correct option is C $10.5\text{C}$


Let us choose an element of length $dx$ at a distance $x$ from the one end of the rod.

Since the element is infinitesimally small, we can assume $\lambda$ to be constant throughout the element.

Infinitesimally small charge $dq$ on the element :

$dq=\lambda dx=(2+x)dx$

The total charge on the rod,

$q=\int dq=\underset{x=0}{\overset{x=3}{\int }}(2+x)dx$

$\Rightarrow q={[2x+\frac{x^2}{2}]}_{x=0}^{x=3}$

$\Rightarrow q=[6+\frac{9}{2}]-\left[0\right]$

$\therefore q=\frac{21}{2}\text{C}=10.5\text{C}$

Hence, option (c) is the correct answer.