Ring of Charge

A ring of charge is a Continuous Charge Distribution that in the shape of a ring. It does not have any area or volume.

Electric Field §

Assume we are looking at the electrical field along the central axis (line that goes through the centre). Then:

Formula §

$$\vec{E}=\frac{k_{e}qz}{(z^2+r^2{)}^{3/2}}\hat{u}$$

• $\vec{E}$ = Electric field of the ring (in $N/C$)
• $q$ = Total charge of ring (in $C$)
• $z$ = Distance from centre. (in $m$)
• $\hat{u}$ = Unit vector perpendicular to the plane of the ring (i.e. direction the ring is ‘facing’ )

Derivation (Without Gauss’ Law) §

Assume we have a circular ring of charge, $S$, with total charge enclosed $q$ and a radius, $r$. Finally, let’s place a reference point charge $A$ ($q_A=+1C)$ along the central axis, at a height, $y$. We can also use $\theta$ (from the Z-Axis) to aid in trigonometry.

We aim to use the formula for Electric Fields, combined with symmetry and integration to obtain the total electric field:

$${\vec{E}}_S=\frac{{\vec{F}}_{SonA}}{q_A}=k_e\times \frac{q_S}{R^2}\times \hat{u}$$

Since we cannot get the exact charge due to the continuous distribution, we can use a very, very small section of the ring, which we can treat as a point charge. To do this, we can use infinitesimals ($\text{d}$).

$$\text{d}{\vec{E}}_S=k_e\times \frac{\text{d}{q}_S}{R^2}\times \hat{u}$$

Now, being a vector, $\hat{u}$ can be divided into it’s X, Y and Z components. However, due to the symmetry of the ring, for every section, $dq$, that we take, there is an opposite section that cancels out the X and Y component:

So the only component we need to worry about the is the Z component:

$${\hat{u}}_Z=\cos (\theta )$$

$$\text{d}{\vec{E}}_S=\text{d}{\vec{E}}_{S_Z}=k_e\times \frac{\text{d}q}{R^2}\times \cos (\theta )$$

Integrate both sides to obtain the total electric field. We can pull out $k_e,R,\theta$ because these stay constant throughout all sections of the circle:

$${\vec{E}}_S=\int \frac{k_e}{R^2}\cos (\theta )\text{d}q=\frac{k_e}{R^2}\cos (\theta )\int 1\text{d}q$$

$${\vec{E}}_S=\frac{k_e}{R^2}\cos (\theta )\times q_S$$

Finally, we can express everything in terms of $z$ and $r$:

$$R^2=z^2+r^2\text{and}\cos (\theta )=\frac{z}{R}=\frac{z}{\sqrt{z^2+r^2}}$$

So our final answer is:

$$\vec{E}=k_e\times \frac{q\times z}{(z^2+r^2{)}^{3/2}}$$