Question

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

• A. $41.3\text{N/C}$
• B. $54.3\text{N/C}$
• C. $31.3\text{N/C}$
• D. $11.3\text{N/C}$

Answer 1

The correct option is B $54.3\text{N/C}$

We know that for a uniformly charged ring, the electric field along the axis at point $\text{P}$ is given by,

$E_x=\frac{kQr}{(r^2+R^2{)}^{3/2}}$

$\Rightarrow E_x=\frac{2\pi k\lambda Rr}{(r^2+R^2{)}^{3/2}}(\because Q=\lambda \times 2\pi R)$

Here,
$r=4\text{m};R=3\text{m};\lambda =+10\text{nC/m}$

So,
$E_x=\frac{2\times 3.14\times 9\times {10}^9\times 10\times {10}^{-9}\times 4\times 3}{(4^2+3^2{)}^{3/2}}$

$\Rightarrow E_x=54.3\text{N/C}$

and, $E_y=0$

$\therefore E_{net}=\sqrt{E_x^2+E_y^2}=\sqrt{{54.3}^2+0^2}=54.3\text{N/C}$

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