Question

For the given uniformly charged ring, magnitude of the net electric field at point $\text{P}$ is (in $\text{N/C}$)

• A. 21
• B. 21.34
• C. 21.3

Answer 1

Answer: (A), (B), (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,
$r=1\text{m};R=5\text{m};\lambda =+10\text{nC/m}$

So,
$E_x=\frac{2\times 3.14\times 9\times {10}^9\times 10\times {10}^{-9}\times 5\times 1}{(1^2+5^2{)}^{3/2}}$

$\Rightarrow E_x=21.3\text{N/C}$

and, $E_y=0$

$\therefore E_{net}=\sqrt{E_x^2+E_y^2}=\sqrt{{21.3}^2+0^2}=21.3\text{N/C}$

Accepted answers: $21,21.3,21.34$