Chapter 25, Problem 66AP

(a)

To determine

The electric potential at the origin due to single charge particle.

Answer to Problem 66AP

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

Explanation of Solution

Write the expression of the electric potential due to single point charge.

    $V=k_{\text{e}}\left(\frac{q}{r}\right)$

Here, $V$ is the electric potential due to single point charge, $k_{\text{e}}$ is the electric constant, $q$ is the charge and $r$ is the distance between the point and charge.

Conclusion:

Substitute $8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2$ for $k_{\text{e}}$, $1.60\text{nC}$ for $q$ and $2.00\text{m}$ for $R$ in the above equation to calculate $V$.

    $\begin{array}{c}V=\left(8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2\right)\left(\frac{1.60\text{nC}}{2.00\text{m}}\right)\\ =\left(8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2\right)\left(\frac{1.60\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}}{2.00\text{m}}\right)\\ =\left(8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2\right)\left(8\times {10}^{-10}\text{C}/\text{m}\right)\\ =7.19\text{V}\end{array}$

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

(b)

To determine

The electric potential at the origin due to two charge particle.

Answer to Problem 66AP

The electric potential at the origin due to two charge particle is $7.67\text{V}$.

Explanation of Solution

Write the expression of the electric potential due to group of  point charge.

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

Here, $V$ is the electric potential due to single point charge, $k_{\text{e}}$ is the electric constant, $q$ is the charge and $r$ is the distance between the point and charge.

For two point chargers the above equation is given as.

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

Here, $q_1$ is the first point charge, $r_1$ is the distance between the $q_1$ charge and given point, $q_2$ is the second point charge and $r_2$ is the distance between the $q_2$ charge and given point.

Conclusion:

Substitute $8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2$ for $k_{\text{e}}$, $0.800\text{nC}$ for $q_1$ and $1.5\text{m}$ for $r_1$, $0.800\text{nC}$ for $q_2$ and $2.5\text{m}$ for $r_2$ in the above equation to calculate $V$.

    $\begin{array}{c}V=\left(8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2\right)\left(\frac{0.800\text{nC}}{1.5\text{m}}+\frac{0.800\text{nC}}{2.5\text{m}}\right)\\ =\left(8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2\right)\left(\frac{0.800\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}}{1.5\text{m}}+\frac{0.800\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}}{2.5\text{m}}\right)\\ =\left(8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2\right)\left(8.53\times {10}^{-10}\text{C}/\text{m}\right)\\ =7.67\text{V}\end{array}$

Therefore, the electric potential at the origin due to two charge particle is $7.67\text{V}$.

(c)

To determine

The electric potential at the origin due to four charge particle.

Answer to Problem 66AP

The electric potential at the origin due to four charge particle is $7.84\text{V}$.

Explanation of Solution

Write the expression of the electric potential due to group of point charge.

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

Here, $V$ is the electric potential due to single point charge, $k_{\text{e}}$ is the electric constant, $q$ is the charge and $r$ is the distance between the point and charge.

For four point chargers the above equation is given as.

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

Here, $q_1$ is the first point charge, $r_1$ is the distance between the $q_1$ charge and given point, $q_2$ is the second point charge, $r_2$ is the distance between the $q_2$ charge and given point, $q_3$ is the third point charge, $r_3$ is the distance between the $q_3$ charge and given point, $q_4$ is the forth point charge and $r_4$ is the distance between the $q_4$ charge and given point.

Conclusion:

Substitute $8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2$ for $k_{\text{e}}$, $0.400\text{nC}$ for $q_1$ and $1.25\text{m}$ for $r_1$, $0.400\text{nC}$ for $q_2$, $1.75\text{m}$ for $r_2$, $0.400\text{nC}$ for $q_3$ and $2.25\text{m}$ for $r_3$, $0.400\text{nC}$ for $q_4$, $2.75\text{m}$ for $r_4$ in the above equation to calculate $V$.

    $\begin{array}{c}V=\left(8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2\right)\left(\frac{0.400\text{nC}}{1.25\text{m}}+\frac{0.400\text{nC}}{1.75\text{m}}+\frac{0.400\text{nC}}{2.25\text{m}}+\frac{0.400\text{nC}}{2.75\text{m}}\right)\\ =\left(8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2\right)\left[\begin{array}{l}\frac{0.400\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}}{1.25\text{m}}+\frac{0.400\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}}{1.75\text{m}}\\ +\frac{0.400\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}}{1.25\text{m}}+\frac{0.400\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}}{1.75\text{m}}\end{array}\right]\\ =\left(8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2\right)\left(8.778\times {10}^{-10}\text{C}/\text{m}\right)\\ =7.84\text{V}\end{array}$

Therefore, the electric potential at the origin due to four charge particle is $7.84\text{V}$.

(d)

To determine

The electric potential given by the exact expression and compare the result.

Answer to Problem 66AP

The electric potential at the origin based on exact expression is $7.899\text{V}$.

Explanation of Solution

Write the given exact expression of electric potential.

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

Here, $l$ is the length of the filament, $a$ is the distance between filament and the origin, $k_{\text{e}}$ is the electric constant and $Q$ is the total charge on the filament.

Conclusion:

Substitute, $8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2$ for $k_{\text{e}}$, $1.60\text{nC}$ for $Q$, $1.00\text{m}$ for $a$ and $2.00\text{m}$ for $l$ in the above equation to calculate $V$.

    $\begin{array}{c}V=\frac{\left(8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2\right)\left(1.60\text{nC}\right)}{2.00\text{m}}\ln \left(\frac{2.00\text{m}+1.00\text{m}}{1.00\text{m}}\right)\\ =\frac{\left(8.988\times {10}^9{\text{N}{\text{.m}}^{\text{2}}/\text{C}}^2\right)\left(1.60\text{nC}\times \frac{1\text{C}}{{10}^9\text{nC}}\right)}{2.00\text{m}}\ln \left(\frac{3.00\text{m}}{1.00\text{m}}\right)\\ =\left(7.19\text{N}\text{.m}/\text{C}\right)\ln \left(\frac{3.00\text{m}}{1.00\text{m}}\right)\\ =7.899\text{V}\end{array}$

Thus, the electric potential based on given expression is $7.899\text{V}$.

In all the values of electric potential in all four parts, the largest value is $7.899\text{V}$.

Therefore, the electric potential at the origin is $7.899\text{V}$.