Chapter 24, Problem 21PQ

Often we have distributions of charge for which integrating to find the electric field may not be possible in practice. In such cases, we may be able to get a good approximate solution by dividing the distribution into small but finite particles and taking the vector sum of the contributions of each. To see how this might work, consider a very thin rod of length L = 16 cm with uniform linear charge density λ = 50.0 nC/m. Estimate the magnitude of the electric field at a point P a distance d = 8.0 cm from the end of the rod by dividing it into n segments of equal length as illustrated in Figure P24.21 for n = 4. Treat each segment as a particle whose distance from point P is measured from its center. Find estimates of E_P for n = 1, 2, 4, and 8 segments.

FIGURE P24.21

To determine

The magnitudes of electric fields at P for the segments $n=1$, $n=2$, $n=3$ and $n=4$.

Answer to Problem 21PQ

The magnitude of electric fields $n=1$ is $2.81\times {10}^3\text{N}/\text{C}$, $n=2$ is $3.40\times {10}^3\text{N}/\text{C}$, $n=4$ is $3.64\times {10}^3\text{N}/\text{C}$ and $n=4$ is $3.72\times {10}^3\text{N}/\text{C}$.

Explanation of Solution

Write the expression to calculate the electric field.

  $E_P=\frac{k{q}_4}{r_1{}^2}+\frac{k{q}_4}{r_2{}^2}+\frac{k{q}_4}{r_3{}^2}+\frac{k{q}_4}{r_4{}^2}$

Here, $E_P$ is the electric field at P, k is the coulomb constant, $r_1$ is the distance from the center of the first segment to P, $r_2$ is the distance from the center of the second segment to P, $r_3$ is the distance from the center of the third segment to P and $r_4$ is the distance from the center of the fourth segment to P and $q_4$ is the one quarter of total charge.

Write the expression to calculate the charge in each segment.

  $q_4=\frac{\lambda L}{4}$

Here, $\lambda$ is the linear charge density and L is the length of the rod

Substitute the above equation in the expression for $E_P$ to rewrite.

  $E_P=\left(\frac{\lambda kL}{4}\right)\left(\frac{1}{r_1{}^2}+\frac{1}{r_2{}^2}+\frac{1}{r_3{}^2}+\frac{1}{r_4{}^2}\right)$                                                                           (I)

Write the expression to calculate $r_1$.

  $r_1=\frac{7L}{8}+d$                                                                                                                       (II)

Here, d is the distance of the point P from the end of the rod.

Write the expression to calculate $r_2$.

  $r_2=\frac{5L}{8}+d$                                                                                                                     (III)

Write the expression to calculate $r_3$.

  $r_3=\frac{3L}{8}+d$                                                                                                                     (IV)

Write the expression to calculate $r_4$.

  $r_4=\frac{L}{8}+d$                                                                                                                     (V)

Substitute the equations (II), (III), (IV) and (V) in (I) to rewrite.

  $E_P=\left(\frac{\lambda kL}{4}\right)\left(\frac{1}{{\left(\frac{7L}{8}+d\right)}^2}+\frac{1}{{\left(\frac{5L}{8}+d\right)}^2}+\frac{1}{{\left(\frac{3L}{8}+d\right)}^2}+\frac{1}{{\left(\frac{L}{8}+d\right)}^2}\right)$

Conclusion:

Substitute $50.0\text{nC}/\text{m}$ for $\lambda$, $8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ for k, $16\text{cm}$ for L and $8\text{cm}$ for d in the above equation to calculate $E_P$.

  $\begin{array}{c}{E}_P=\left(\frac{\left(50.0\text{nC}/\text{m}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\left(16\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{4}\right)\\ \left(\begin{array}{l}\frac{1}{{\left(\frac{7\left(\left(16\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\right)}{8}+\left(8\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\right)}^2}+\frac{1}{{\left(\frac{5\left(\left(16\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\right)}{8}+\left(8\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\right)}^2}+\\ \frac{1}{{\left(\frac{3\left(\left(16\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\right)}{8}+\left(8\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\right)}^2}+\frac{1}{{\left(\frac{\left(16\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}{8}+\left(8\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\right)}^2}\end{array}\right)\\ \approx 3.64\times {10}^3\text{N}/\text{C}\end{array}$

Similarly, by following the same concepts the electric field for $n=1$ is $2.81\times {10}^3\text{N}/\text{C}$, $n=2$ is $3.40\times {10}^3\text{N}/\text{C}$, $n=4$ is $3.64\times {10}^3\text{N}/\text{C}$ and $n=4$ is $3.72\times {10}^3\text{N}/\text{C}$.

Therefore, the magnitude of electric fields $n=1$ is $2.81\times {10}^3\text{N}/\text{C}$, $n=2$ is $3.40\times {10}^3\text{N}/\text{C}$, $n=4$ is $3.64\times {10}^3\text{N}/\text{C}$ and $n=4$ is $3.72\times {10}^3\text{N}/\text{C}$.