Question

A uniformly charged rod of length $4\text{cm}$ and linear charge density $\lambda =30\mu \text{C/m}$ is placed as shown in figure. Calculate the $x-$component of electric field at point $\text{P}$.

• A. $36\times {10}^5\text{N/C}$
• B. $9\times {10}^5\text{N/C}$
• C. $1.8\times {10}^3\text{N/C}$
• D. $27\times {10}^5\text{N/C}$

Answer 1

The correct option is A $36\times {10}^5\text{N/C}$


From above diagram,

${\theta }_1={53}^{\circ }\&{\theta }_2=0^{\circ }$

$r_1=3\text{cm}=0.03\text{m}$

$x-$ component or parallel component of field due to uniformly distributed line charge is given by

$E_{||}=\frac{k\lambda }{r_1}(\cos {\theta }_2-\cos {\theta }_1)$

$\Rightarrow E_{||}=\frac{9\times {10}^9\times 30\times {10}^{-6}}{3\times {10}^{-2}}(\cos 0^{\circ }-\cos {53}^{\circ })$

$\Rightarrow E_{||}=\frac{9\times 30}{3}\times {10}^5\times \frac{2}{5}=36\times {10}^5\text{N/C}$

Hence, option $\left(a\right)$ is correct.