Solved Problem on Coulomb's Law and Electric Field

Two concentric rings are located on the same plane. The ring of radius R₁ has a charge Q₁, and the ring of radius R₂ has a charge Q₂. The electric field vector produced by a ring of radius r at a distance z from the center is given by

\[ \begin{gather} \mathbf{E}=\frac{1}{4\pi \epsilon_0}\frac{Qz}{\left(r^2+z^2\right)^{3/2}}\;\mathbf{k} \end{gather} \]

Determine the electric field vector:
a) In the common center of the two rings;
b) At a point located at a distance z, much greater than R₁ and R₂.

Problem data:

Radius of ring 1: R₁;
Charge of ring 1: Q₁;
Radius of ring 2: R₂;
Charge of ring 2: Q₂.

Problem diagram:

We choose a frame of reference at the center of the rings with R₁>R₂ (Figure 1-A).

Figure 1

We want to calculate the electric field at the common center of the two rings, at the origin at z=0, and any point z at a distance much greater than the radii R₁ and R₂ of the rings (Figure 1-B).

Solution

The total electric field vector is given by the sum of the electric field vectors produced by each of the rings

\[ \begin{gather} \bbox[#99CCFF,10px] {\mathbf{E}=\mathbf{E}_1+\mathbf{E}_2} \end{gather} \]

a) For ring 1, the electric field vector will be

\[ \begin{gather} \mathbf{E}_1=\frac{1}{4\pi \epsilon_0}\frac{Q_1\times0}{\left(R_1^2+0^2\right)^{3/2}}\;\mathbf{k}=0 \end{gather} \]

For ring 2, the electric field vector will be

\[ \begin{gather} \mathbf{E}_2=\frac{1}{4\pi \epsilon_0}\frac{Q_1\times0}{\left(R_2^2+0^2\right)^{3/2}}\;\mathbf{k}=0 \end{gather} \]

The total electric field vector will be

\[ \begin{gather} \mathbf{E}=0+0 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathbf{E}=0} \end{gather} \]

b) For ring 1, the electric field vector at a distance z will be

\[ \begin{gather} \mathbf{E}_1=\frac{1}{4\pi \epsilon_0}\frac{Q_1z}{\left(R_1^2+z^2\right)^{3/2}}\;\mathbf{k} \end{gather} \]

if z≫R₁, we can neglect the value of R₁

\[ \begin{gather} \mathbf{E}_1=\frac{1}{4\pi \epsilon_0}\frac{Q_1z}{\left(z^2\right)^{3/2}}\;\mathbf{k}\\[5pt] \mathbf{E}_1=\frac{1}{4\pi\epsilon_{0}}\frac{Q_1z}{z^{2\times{\frac{3}{2}}}}\;\mathbf{k}\\[5pt] \mathbf{E}_1=\frac{1}{4\pi\epsilon_0}\frac{Q_1\cancel z}{z^{\cancelto{2}{3}}}\;\mathbf{k}\\[5pt] \mathbf{E}_1=\frac{1}{4\pi\epsilon_0}\frac{Q_1}{z^2}\;\mathbf{k} \end{gather} \]

For ring 2, the electric field vector will be

\[ \begin{gather} \mathbf{E}_2=\frac{1}{4\pi \epsilon_0}\frac{Q_1z}{\left(R_2^2+z^2\right)^{3/2}}\;\mathbf{k} \end{gather} \]

if z≫R₂, we can neglect the value of R₂

\[ \begin{gather} \mathbf{E}_2=\frac{1}{4\pi \epsilon_0}\frac{Q_2z}{\left(z^2\right)^{3/2}}\;\mathbf{k}\\[5pt] \mathbf{E}_2=\frac{1}{4\pi\epsilon_{0}}\frac{Q_2z}{z^{2\times{\frac{3}{2}}}}\;\mathbf{k}\\[5pt] \mathbf{E}_2=\frac{1}{4\pi\epsilon_0}\frac{Q_2\cancel z}{z^{\cancelto{2}{3}}}\;\mathbf{k}\\[5pt] \mathbf{E}_2=\frac{1}{4\pi\epsilon_0}\frac{Q_2}{z^2}\;\mathbf{k} \end{gather} \]

The total electric field vector will be

\[ \begin{gather} \mathbf{E}=\frac{1}{4\pi \epsilon_0}\frac{Q_1}{z^2}\;\mathbf{k}+\frac{1}{4\pi\epsilon_{0}}\frac{Q_2}{z^2}\;\mathbf{k} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathbf{E}=\frac{1}{4\pi \epsilon_0}\frac{\left(Q_1+Q_2\right)}{z^2}\;\mathbf{k}} \end{gather} \]

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