A uniform charge of 3.14 × 10^ -6 C is distributed over a circular...

The electric field $E$ at a distance $x$ on the axis of a uniformly charged ring is given by:

$E = \dfrac{1}{4 \pi \epsilon_0} \dfrac{q x}{(R^2 + x^2)^{3/2}}$

The magnetic field $B$ at a distance $x$ on the axis of a circular current loop is:

$B = \dfrac{\mu_0 i R^2}{2 (R^2 + x^2)^{3/2}}$

The equivalent current $i$ due to the rotating charge is:

$i = \dfrac{q}{T} = \dfrac{q \omega}{2 \pi}$

Substituting $i$ into the expression for $B$:

$B = \dfrac{\mu_0 q \omega R^2}{4 \pi (R^2 + x^2)^{3/2}}$

Taking the ratio of $E$ to $B$:

$\dfrac{E}{B} = \dfrac{\dfrac{1}{4 \pi \epsilon_0} \dfrac{q x}{(R^2 + x^2)^{3/2}}}{\dfrac{\mu_0 q \omega R^2}{4 \pi (R^2 + x^2)^{3/2}}} = \dfrac{x}{\epsilon_0 \mu_0 \omega R^2}$

Using the relation $c = \dfrac{1}{\sqrt{\epsilon_0 \mu_0}}$, we get:

$\dfrac{E}{B} = \dfrac{x c^2}{\omega R^2}$

Given values are:
$x = 5.00 \text{ cm} = 0.05 \text{ m}$
$R = 20.0 \text{ cm} = 0.2 \text{ m}$
$\omega = 60.0 \text{ rad s}^{-1}$
$c = 3 \times 10^8 \text{ m s}^{-1}$

Substituting the values:

$\dfrac{E}{B} = \dfrac{0.05 \times (3 \times 10^8)^2}{60 \times (0.2)^2}$

$\dfrac{E}{B} = \dfrac{0.05 \times 9 \times 10^{16}}{60 \times 0.04}$

$\dfrac{E}{B} = \dfrac{0.45 \times 10^{16}}{2.4}$

$\dfrac{E}{B} = \dfrac{4.5}{24} \times 10^{16}$

$\dfrac{E}{B} = \dfrac{3}{16} \times 10^{16}$

$\dfrac{E}{B} = 0.1875 \times 10^{16} = 1.875 \times 10^{15} \text{ m s}^{-1}$