An arc of a circle of radius a and central angle θ₀ carries an electric charge Q uniformly distributed along the arc. Determine:
a) The electric field vector, at the points of the line passing through the center of the arc, and is perpendicular to the plane containing the arc;
b) The electric field vector in the center of curvature of the arc;
c) The electric field vector when the central angle tends to zero.

Problem data:

• Radius of the arc:    a;
• Central angle of the arc:    θ₀;
• Electric charge on the arc:    Q.

Problem diagram:

The position vector r goes from an element of charge dq to point P, where we want to calculate the electric field, the vector r_q locates the charge element relative to the origin of the reference frame, and the vector r_p locates point P (Figure 1-A).

Figure 1

From the geometry of the problem, we choose cylindrical coordinates (Figure 1-B). The r_q vector, which is on the yz plane, is written as \( \mathbf r_q=y\;\mathbf j+z\;\mathbf k \) and the r_p vector only has a component in the i direction, \( \mathbf r_p=x\;\mathbf i \) (contrary to what is usual, where the r_q vector is on the xy plane and the axis of the cylinder is in the k direction), the position vector will be

From equation (I), the magnitude of the position vector will be

where x, y, and z, in cylindrical coordinates, are given by

Solution:

a) The electric field vector is given by

Using the equation for the linear density charge λ, we have the charge element dq.

where ds is an arc element with angle dθ (Figure 2).

\[ \begin{gather} ds=a\;d\theta \tag{VI} \end{gather} \]

substituting the equation (VI) into equation (V).

\[ \begin{gather} dq=\lambda a\;d\theta \tag{VII} \end{gather} \]

Figure 2

Substituting equations (I), (II), and (VII) into equation (IV).

substituting equations (III) into equation (VIII).

As the charge density λ and the radius a are constants, they are moved outside the integral, and the integral of the sum is equal to the sum of the integrals.

As there is symmetry, we can divide the central angle θ₀ into two parts measuring \( \frac{\theta_0}{2} \) clockwise and \( -{\frac{\theta_0}{2}} \) counterclockwise (Figure 3), the integration limits will be \( -{\frac{\theta_0}{2}} \) and \( \frac{\theta_0}{2} \).

\[ \begin{gather} \mathbf E=\frac{1}{4\pi\epsilon_0}\frac{\lambda a}{\left(x^2+a^2\right)^{3/2}}\left(x\int_{-{\frac{\theta_0}{2}}}^{{\frac{\theta_0}{2}}}d\theta\;\mathbf i-a\int_{-{\frac{\theta_0}{2}}}^{{\frac{\theta_0}{2}}}\cos\theta\;d\theta\;\mathbf j-a\int_{-{\frac{\theta_0}{2}}}^{{\frac{\theta_0}{2}}}\sin\theta\;d\theta\;\mathbf k\right) \end{gather} \]

Figure 3

Note: The integral in the k direction is equal to zero, because an element of charge dq, produces at a point, an element of the field that can be decomposed into the elements, dE_x, −dE_y, and −dE_z (Figure 4-A). Another element of charge placed in a symmetrical position produces at the same point another element of the field that can be decomposed into the elements dE_x, −dE_y, and dE_z (Figure 4-B). Hence, the elements in the k direction cancel, and only elements in the i and j directions contribute to the total field.

Figure 4

The total charge of the arc is Q, and its length is aθ₀, so the linear density of charge can be written

substituting equation (X) into equation (IX).

\[ \begin{gather} \mathbf E=\frac{1}{4\pi\epsilon_0}\frac{Q}{a\theta_0}\frac{a}{\left(x^2+a^2\right)^{3/2}}\left(x\theta_0\;\mathbf i-2a\sin\frac{\theta_0}{2}\;\mathbf j\right)\\[5pt] \mathbf E=\frac{1}{4\pi\epsilon_0}\frac{Q}{\left(x^2+a^2\right)^{3/2}}\left(\frac{1}{\theta_0}x\theta_0\;\mathbf i-\frac{1}{\theta_0}2a\sin\frac{\theta_0}{2}\;\mathbf j\right) \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathbf E=\frac{1}{4\pi\epsilon_0}\frac{Q}{\left(x^2+a^2\right)^{3/2}}\left(x\;\mathbf i-\frac{2a}{\theta_0}\sin\frac{\theta_0}{2}\;\mathbf j\right)} \end{gather} \]

Figure 5

b) In the center of curvature, we have x = 0, substituting this in the solution of the previous item (Figure 6).

\[ \begin{gather} \mathbf E=\frac{1}{4\pi\epsilon_0}\frac{Q}{\left(0^2+a^2\right)^{3/2}}\left(0\;\mathbf i-\frac{2a}{\theta_0}\sin\frac{\theta_0}{2}\;\mathbf j\right)\\[5pt] \mathbf E=\frac{-{1}}{4\pi\epsilon_0}\frac{Q}{a^{3}}\frac{2a}{\theta_0}\sin\frac{\theta_0}{2}\;\mathbf j \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {\mathbf E=\frac{-{1}}{4\pi\epsilon_0}\frac{Q}{a^2}\frac{2}{\theta_0}\sin\frac{\theta_0}{2}\;\mathbf j} \end{gather} \]

Figure 6

c) When the central angle tends to zero (\( \theta_0\rightarrow 0 \)), the arc tends to a point charge, applying the limit to the solution of the previous item (Figure 7).

\[ \begin{gather} \mathbf E=\underset{\theta_0\rightarrow0}{\lim}-\frac{1}{4\pi\epsilon_0}\;\frac{Q}{a^2}\frac{2}{\theta_0}\sin\frac{\theta_0}{2}\;\mathbf j \end{gather} \]

turning the term \( \frac{2}{\theta_0} \) upside down.

\[ \begin{gather} \mathbf E=\underset{\theta_0\rightarrow 0}{\lim}-{\frac{1}{4\pi\epsilon_0}\frac{Q}{a^2}\dfrac{\sin\dfrac{\theta_0}{2}}{\dfrac{\theta_0}{2}}\;\mathbf j} \end{gather} \]

Figure 7

and the result is reduced to the electric field vector of a point charge.