Question

For the given uniformly charged ring, magnitude of net electric field at point $\text{P}$ is

• A. $5.6\text{mN/C}$
• B. $4.6\text{mN/C}$
• C. $3.6\text{mN/C}$
• D. $2.6\text{mN/C}$

Answer 1

The correct option is A $5.6\text{mN/C}$

Given:

$\lambda =+10\text{nC/m};R=1\text{mm}=1\times {10}^{-3}\text{m};r=10\text{m}$

From the above data, we observe that $r>>R$.

We know that, when $r>>R$, the ring acts like a point charge.

So, Electric field at point $\text{P}$ along $x-$ direction,

$E_x=\frac{kQ}{r^2}=\frac{2\pi k\lambda R}{r^2}(\because Q=\lambda \times 2\pi R)$

$\Rightarrow E_x=\frac{2\times 3.14\times 9\times {10}^9\times 10\times {10}^{-9}\times 1\times {10}^{-3}}{{10}^2}$

$\Rightarrow E_x=5.6\times {10}^{-3}\text{N/C}$

$\Rightarrow E_x=5.6\text{mN/C}$

and, $E_y=0$

$\therefore E_{net}=\sqrt{E_x^2+E_y^2}=\sqrt{{5.6}^2+0^2}=5.6\text{mN/C}$

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