A thin, $0.1 \mathrm{~m}$ radius coil has an evenly distributed electric charge. What is this charge if the maximum electric field strength along the symmetry axis of the coil is $8 \cdot 10^{4} \sqrt{3} \mathrm{~V} / \mathrm{m}$? What is the electric field strength at a distance of $0.2 \mathrm{~m}$ from the center?
Divide the ring into tiny pieces and denote the charge on each elementary piece by $\Delta Q$ (see the figure)!

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The total charge on the ring is clearly $Q=\Sigma \Delta Q$. By symmetry, the electric field along the axis of the ring (which is perpendicular to the plane of the ring) can only be directed along the axis. Calculate the contribution of a charge $\Delta Q$ to the component of the electric field along this axis at a distance $x$ from the center! The magnitude of the electric field is $|\Delta E|=k \cdot \Delta Q /\left(x^{2}+r^{2}\right)$, and $\cos \alpha=x / \sqrt{x^{2}+r^{2}}$, so

$$
\Delta E_{x}=k \cdot \Delta Q \cdot \frac{x}{\left(x^{2}+r^{2}\right)^{3 / 2}}
$$

The total electric field is

$$
E(x)=\Sigma \Delta E_{x}=k \frac{x}{\left(x^{2}+r^{2}\right)^{3 / 2}} \cdot \Sigma \Delta Q=k Q \cdot \frac{x}{\left(x^{2}+r^{2}\right)^{3 / 2}}
$$

This function has a maximum at $x=x_{0}$ where the expression $x /\left(x^{2}+r^{2}\right)^{3 / 2}$ takes its maximum value. This can be determined graphically, numerically using a computer, or by differential calculus; the result is $x_{0}=r / \sqrt{2}$. Since $E\left(x_{0}\right)=E_{\max }$ is known, the magnitude of the charge can be calculated, and $Q=4 \cdot 10^{-7} \mathrm{C}$ is obtained. At a distance of $x_{1}=0.2 \mathrm{~m}$ from the center, the electric field is $E\left(x_{1}\right)=6.44 \cdot 10^{4} \mathrm{~V} / \mathrm{m}$.