Chapter 25, Problem 25.66AP

A uniformly charged filament lies along the x axis between x = a = 1.00 m and x = a + ℓ = 3.00 m as shown in Figure P25.66. The total charge on the filament is 1.60 nC. Calculate successive approximations for the electric potential at the origin by modeling the filament as (a) a single charged particle at x = 2.00 m, (b) two 0.800-nC charged particles at x = 1.5 m and x = 2.5 m, and (c) four 0.400-nC charged particles at x = 1.25 m, x = 1.75 m, x = 2.25 m, and x = 2.75 m. (d) Explain how the results compare with the potential given by the exact expression

$\text{v}=\frac{k_{\mathcal{l}}Q}{\mathcal{l}}\ln \left(\frac{\mathcal{l}+a}{a}\right)$

To determine

The electric potential due to single charge particle.

Answer to Problem 25.66AP

The electric potential due to single charge particle is $7.19\text{V}$.

Explanation of Solution

Given info: The total charge on the filament is $1.60\text{nC}$ and the distance of the single charged particle from $\text{y}$-axis is $2.00\text{m}$.

Formula to calculate the electric potential due to point change is,

$V=k_{\text{e}}\frac{q}{r}$

Here,

$V$ is the electric potential.

$k_{\text{e}}$ is the coulomb’s constant.

$q$ is the charge on the single particle.

Substitute $8.99\times {10}^9{\text{N}\cdot {\text{m}}^2/\text{C}}^2$ for $k_{\text{e}}$, $1.60\text{nC}$ for $q$ and $2.00\text{m}$ for $r$ in above equation to find $V$.

$\begin{array}{c}V=\left(8.99\times {10}^9{\text{N}\cdot {\text{m}}^2/\text{C}}^2\right)\frac{\left(1.60\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)}{2.00\text{m}}\\ V=7.19\text{V}\end{array}$

Conclusion:

Therefore, the electric potential due to single charge particle is $7.19\text{V}$.

To determine

The electric potential due to two charged particles.

Answer to Problem 25.66AP

The electric potential due to two charged particles is $7.67\text{V}$.

Explanation of Solution

Given info: The charge on two particles is $0.800\text{nC}$, the distance of first particle from $\text{y}$-axis is $1.5\text{m}$ and the distance of second particle from $\text{y}$-axis is $2.5\text{m}$.

Formula to calculate the total electric potential tow charge particles is,

$V=k_{\text{e}}\left(\frac{q_1}{r_1}+\frac{q_2}{r_2}\right)$

Here,

$q_1$ is the charge on first particle.

$q_2$ is the charge on second particle.

$r_1$ is distance of first particle from $\text{y}$-axis.

$r_2$ is the distance of second particle from $\text{y}$-axis.

Substitute $8.99\times {10}^9{\text{N}\cdot {\text{m}}^2/\text{C}}^2$ for $k_{\text{e}}$, $0.800\text{nC}$ for $q_1$, $0.800\text{nC}$ for $q_2$, $1.5\text{m}$ for $r_1$ and $2.5\text{m}$ for $r_2$ in above equation to find $V$.

$\begin{array}{c}V={\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/\text{C}\right)}^2\left(\frac{\left(0.800\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)}{1.5\text{m}}+\frac{\left(0.800\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)}{2.5\text{m}}\right)\\ =7.67\text{V}\end{array}$

Conclusion:

Therefore, the electric potential due to two charged particles is $7.67\text{V}$.

To determine

The electric potential due to four charged particles.

Answer to Problem 25.66AP

The electric potential due to four charged particles is $7.84\text{V}$.

Explanation of Solution

Given info: The charge on four particles is $0.400\text{nC}$, the distance of first particle from $\text{y}$-axis is $1.25\text{m}$, the distance of second particle from $\text{y}$-axis is $1.75\text{m}$, the distance of third particle from $\text{y}$-axis is $2.25\text{m}$ and the distance of fourth particle from $\text{y}$-axis is $2.75\text{m}$.

Formula to calculate the total electric potential tow charge particles is,

$V=k_{\text{e}}\left(\frac{q_1\text{'}}{r_1}+\frac{q_2\text{'}}{r_2}+\frac{q_3}{r_3}+\frac{q_4}{r_4}\right)$

Here,

$q_1\text{'}$ is the charge on first particle.

$q_2\text{'}$ is the charge on second particle.

$q_3$ is the charge on third particle.

$q_4$ is the charge on fourth particle.

$r_1$ is distance of first particle from $\text{y}$-axis.

$r_2$ is the distance of second particle from $\text{y}$-axis.

$r_3$ is the distance of third particle from $\text{y}$-axis.

$r_4$ is the distance of fourth particle from $\text{y}$-axis.

Substitute $8.99\times {10}^9{\text{N}\cdot {\text{m}}^2/\text{C}}^2$ for $k_{\text{e}}$, $0.400\text{nC}$ for $q_1$, $0.400\text{nC}$ for $q_2$, $0.400\text{nC}$ for $q_3$, $0.400\text{nC}$ for $q_4$, $1.25\text{m}$ for $r_1$ and $1.75\text{m}$ for $r_2$, $2.25\text{m}$ for $r_3$ and $2.75\text{m}$ for $r_4$ in above equation to find $V$.

$\begin{array}{c}V={\left(8.99\times {10}^9\text{N}\cdot {\text{m}}^2/\text{C}\right)}^2\left(\begin{array}{l}\frac{\left(0.400\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)}{1.25\text{m}}+\frac{\left(0.400\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)}{1.75\text{m}}\\ +\frac{\left(0.400\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)}{2.25\text{m}}+\frac{\left(0.400\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)}{2.75\text{m}}\end{array}\right)\\ =7.84\text{V}\end{array}$

Conclusion:

Therefore, the electric potential due to four charged particles is $7.84\text{V}$.

To determine

The comparison between the electric potential due to charged filament and the calculated value of electric potential in part (a), part (b) and part (c).

Answer to Problem 25.66AP

The calculated value of electric potential due to charged filament is $7.84\text{V}$ which is greater than the calculated value of electric potential in part (a), part (b) and part (c).

Explanation of Solution

The total charge on the filament is $1.60\text{nC}$, the distance of the first end of the filament from $\text{y}$-axis is $1.00\text{m}$ and the distance of the last end of the filament from $\text{y}$-axis is $3.00\text{m}$.

The given expression of the electric potential is,

$V=\frac{K_{e}Q}{l}\left(\frac{l+a}{a}\right)$

Here,

$a$ is the distance of particle from $\text{y}$-axis.

$l$ is the length of the particle.

Substitute $8.99\times {10}^9{\text{N}\cdot {\text{m}}^2/\text{C}}^2$ for $k_{\text{e}}$, $1.60\text{nC}$ for $Q$, $1.00\text{m}$ for $l$ and $3.00\text{m}$ for $l+a$ in above equation to find $V$.

$\begin{array}{c}V=\frac{\left(8.99\times {10}^9{\text{N}\cdot {\text{m}}^2/\text{C}}^2\right)\left(1.60\text{nC}\left(\frac{{10}^{-9}\text{C}}{1\text{nC}}\right)\right)}{2.00\text{m}}\left(\frac{3.00\text{m}}{1.00\text{m}}\right)\\ =7.899\text{V}\end{array}$

The electric potential due to charged filament is $7.84\text{V}$ which is greater than the calculated value of electric potential in part (a), part (b) and part (c).

Conclusion:

Therefore, the calculated value of electric potential is $7.84\text{V}$ which is greater than the calculated value of electric potential in part (a), part (b) and part (c).