**A ring-shaped conductor with radius a carries a total charge Q uniformly distributed around it. A point P lies a distance x on an axis through the centre of the ring-shaped conductor. Find the electric field at P. (Note: Symmetry in the problem)**

Since the problem states that the charge is uniformly distributed, the linear charge density, $\lambda$ is:

$$\lambda = \frac{Q}{2 \pi a}$$

We will now find the electric field at P due to a “small” element of the ring of charge. Let $dS$ be the small element. Note that $dS = a \, d \theta$ as $dS$ is just the arc length (Recall: arc length = radius X angle ). Hence,

$$\begin{aligned} dQ &= \lambda \, dS \\ &= \lambda a \, d \theta \\ &= \frac{Q}{2 \pi } \, d \theta \end{aligned}$$

From the symmetry of the problem, we note that only the horizontal component of the electric field will survive at P. The vertical components cancel one another out as you trace out a full circle. Finding the horizontal component:

$$\begin{aligned} dE_{x} &= dE \, \text{cos} \, \alpha \\ &= \frac{dQ}{4 \pi \epsilon_{0} (x^{2} + a^{2})} \, \text{cos} \, \alpha \end{aligned}$$

To get rid of the $\text{cos} \, \alpha$, we note that $\text{cos} \, \alpha = \frac{x}{r} = \frac{x}{ \sqrt{x^{2} + a^{2}}}$ as seen from the diagram. Hence,

$$\begin{aligned} dE_{x} &= \frac{x dQ}{4 \pi \epsilon_{0} (x^{2} + a^{2})^{\frac{3}{2}} } \\ &= \frac{xQ}{8 \pi^{2} \epsilon_{0} (x^{2} + a^{2} )^{\frac{3}{2}}} \, d \theta \end{aligned}$$

Finally, after integrating the above equation from 0 to $2 \pi$ (which is just multiplying by $2 \pi$),

$$\vec{E} = \frac{xQ}{4 \pi \epsilon_{0} \left( x^{2} + a^{2} \right)^{\frac{3}{2}}} \, \hat{i}$$

Next: Electric Field Of A Line Of Charge