Question

A $10\text{cm}$ long rod carries a charge of $+50\mu \text{C}$ distributed uniformly along its length. Find the magnitude of the electric field at a point $10\text{cm}$ from both the ends of the rod.

• A. $4.2\times {10}^7\text{N/C}$
• B. $5.2\times {10}^7\text{N/C}$
• C. $6.2\times {10}^7\text{N/C}$
• D. $7.2\times {10}^7\text{N/C}$

Answer 1

The correct option is B $5.2\times {10}^7\text{N/C}$

Given:
$L=10\text{cm}=10\times {10}^{-2}\text{m}$
$Q=+50\mu \text{C}=+50\times {10}^{-6}\text{C}$

$\therefore \lambda =\frac{Q}{L}=\frac{50\times {10}^{-6}}{10\times {10}^{-2}}=5\times {10}^{-4}\text{C/m}$

For the point to be at $10\text{cm}$ from both the ends of the rod, the point must be on the equatorial line.


From the figure,

$\alpha =\beta =\theta \left(\text{let}\right)={30}^{\circ }$

and, $r=10\cos {30}^{\circ }=5\sqrt{3}\text{cm}=5\sqrt{3}\times {10}^{-2}\text{m}$

From formula, electric field at the equatorial point $\text{P}$

$E_P=\frac{2kQ\sin \theta }{Lr}=\frac{2k\lambda \sin \theta }{r}$

$\Rightarrow E_P=\frac{2\times 9\times {10}^9\times 5\times {10}^{-4}\times \sin {30}^{\circ }}{5\sqrt{3}\times {10}^{-2}}$

$\Rightarrow E_P=5.2\times {10}^7\text{N/C}$

Hence, option $\left(b\right)$ is the correct answer.