A thin ring of radius 15 cm carries a total charge of 20 μC. What is the electric field at a point on the ring’s axis 30 cm from the center?

What we are finding on a ring’s axis

A uniformly charged ring is symmetric, so at any point on its axis, the sideways (radial) components of the field from opposite parts of the ring cancel. Only the axial components add up, giving a standard formula for $E$ on the axis.

Using the on-axis ring formula

Let the ring radius be $a$ and the axial distance from the center be $z$. The electric field magnitude on the axis is $$ E=\frac{1}{4\pi\varepsilon_0}\frac{Qz}{(z^2+a^2)^{3/2}}. $$ Here,

• $a=15\,\text{cm}=0.15\,\text{m}$
• $z=30\,\text{cm}=0.30\,\text{m}$
• $Q=20\,\mu\text{C}=20\times 10^{-6}\,\text{C}$
• $k=\frac{1}{4\pi\varepsilon_0}=8.99\times 10^9\,\text{N}\,\text{m}^2/\text{C}^2$

Plugging in the numbers carefully

Compute the geometry term: $$ z^2+a^2=(0.30)^2+(0.15)^2=0.09+0.0225=0.1125 $$ $$ (z^2+a^2)^{3/2}=(0.1125)^{3/2}\approx 0.0377 $$ Now the numerator: $$ kQz=(8.99\times 10^9)(20\times 10^{-6})(0.30)\approx 5.39\times 10^4 $$ So, $$ E=\frac{5.39\times 10^4}{0.0377}\approx 1.43\times 10^6\ \text{N/C}. $$

Direction of the field

Because $Q$ is positive, the electric field points away from the ring. At a point on the axis on one side of the ring, that direction is along the axis away from the center (on that side).