Question

For the given uniformly charged ring, what will be the maximum value of electric field?

• A. $116\text{N/C}$
• B. $126\text{N/C}$
• C. $136\text{N/C}$
• D. $146\text{N/C}$

Answer 1

The correct option is B $126\text{N/C}$

Given:
$R=3\sqrt{3}\text{m};\lambda =+30\text{nC/m}=+30\times {10}^{-9}\text{C/m}$

We know that, maximum value of electric field for a ring of uniformly distributed charge over its circumference is given by

$E_{\text{max}}=\frac{2kQ}{3\sqrt{3}{R}^2}=\frac{4\pi k\lambda }{3\sqrt{3}R}$ $(\because Q=\lambda \times 2\pi r)$

$\Rightarrow E_{\text{max}}=\frac{4\times 3.14\times 9\times {10}^9\times 30\times {10}^{-9}}{3\sqrt{3}\times 3\sqrt{3}}$

$\Rightarrow E_{\text{max}}\approx 125.6=126\text{N/C}$

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

Why this question? Tip: Always remember that in this case electric field is maximum at a distance of $R/\sqrt{2}$ from the centre of the ring.