Chapter 20, Problem 44A

To determine

To rank:The pairs of point charges according to the magnitude of electrostatic force.

Answer to Problem 44A

The rank of the pairs of point charges according to the magnitude of electrostatic force is,

E. The force of a $1.0\mu \text{C}$ charge and a $2.5\mu \text{C}$ charge is $F_{\text{E}}=2.25\text{ N}$ .

A. The force of two $7.0\text{ nC}$ charges is $F_{\text{A}}=1.10\times {10}^{-5}\text{ N}$ .

B. The force of two $5.0\text{ nC}$ charges is $F_{\text{B}}=5.6\times {10}^{-6}\text{ N}$ .

C. The force of two $2.5\text{ nC}$ charges is $F_{\text{C}}=5.6\times {10}^{-6}\text{ N}$ .

D. The force of a $2.5\text{ nC}$ charge and a $5.0\text{ nC}$ charge is $F_{\text{D}}=2.8\times {10}^{-6}\text{ N}$ .

The ties are B and C.

Explanation of Solution

Given:

The pairs of point charges are,

A. two $7.0\text{ nC}$ charges separated by $0.20\text{ m}$ .

B. two $5.0\text{ nC}$ charges separated by $0.20\text{ m}$ .

C. two $2.5\text{ nC}$ charges separated by $0.10\text{ m}$ .

D. a $2.5\text{ nC}$ charge and a $5.0\text{ nC}$ charge separated by $0.20\text{ m}$ .

E. $1.0\mu \text{C}$ charge and a $2.5\mu \text{C}$ charge separated by $0.10\text{ m}$ .

Formula used:

The expression for Coulomb’s law is,

$F=K\frac{q_{\text{A}}{q}_{\text{B}}}{r^2}$

Here, $K$ is the Coulomb’s constant, $q_{\text{A}}\text{ and }{q}_{\text{B}}$ are the two charges, and $r$ is the distance between the charges.

Calculation:

Consider the Coulomb’s constant as $K=9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$ .

The force of two $7.0\text{ nC}$ charges is,

$\begin{array}{l}{F}_{\text{A}}=K\frac{q_{\text{A}}{q}_{\text{B}}}{r^2}\\ F_{\text{A}}=\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times \frac{\left(\text{7}\text{.0 nC}\times \frac{{10}^{-9}\text{ C}}{1\text{ nC}}\right)\left(\text{7}\text{.0 nC}\times \frac{{10}^{-9}\text{ C}}{1\text{ nC}}\right)}{{\left(0.20\text{ m}\right)}^2}\\ F_{\text{A}}=1.10\times {10}^{-5}\text{ N}\end{array}$

The force of two $5.0\text{ nC}$ charges is,

$\begin{array}{l}{F}_{\text{B}}=K\frac{q_{\text{A}}{q}_{\text{B}}}{r^2}\\ F_{\text{B}}=\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times \frac{\left(\text{5}\text{.0 nC}\times \frac{{10}^{-9}\text{ C}}{1\text{ nC}}\right)\left(\text{5}\text{.0 nC}\times \frac{{10}^{-9}\text{ C}}{1\text{ nC}}\right)}{{\left(0.20\text{ m}\right)}^2}\\ F_{\text{B}}=5.6\times {10}^{-6}\text{ N}\end{array}$

The force of two $2.5\text{ nC}$ charges is,

$\begin{array}{l}{F}_{\text{C}}=K\frac{q_{\text{A}}{q}_{\text{B}}}{r^2}\\ F_{\text{C}}=\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times \frac{\left(\text{2}\text{.5 nC}\times \frac{{10}^{-9}\text{ C}}{1\text{ nC}}\right)\left(\text{2}\text{.5 nC}\times \frac{{10}^{-9}\text{ C}}{1\text{ nC}}\right)}{{\left(0.10\text{ m}\right)}^2}\\ F_{\text{C}}=5.6\times {10}^{-6}\text{ N}\end{array}$

The force of a $2.5\text{ nC}$ charge and a $5.0\text{ nC}$ charge is,

$\begin{array}{l}{F}_{\text{D}}=K\frac{q_{\text{A}}{q}_{\text{B}}}{r^2}\\ F_{\text{D}}=\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times \frac{\left(\text{2}\text{.5 nC}\times \frac{{10}^{-9}\text{ C}}{1\text{ nC}}\right)\left(\text{5}\text{.0 nC}\times \frac{{10}^{-9}\text{ C}}{1\text{ nC}}\right)}{{\left(0.20\text{ m}\right)}^2}\\ F_{\text{D}}=2.8\times {10}^{-6}\text{ N}\end{array}$

The force of a $1.0\mu \text{C}$ charge and a $2.5\mu \text{C}$ charge is,

$\begin{array}{l}{F}_{\text{E}}=K\frac{q_{\text{A}}{q}_{\text{B}}}{r^2}\\ F_{\text{E}}=\left(9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}\right)\times \frac{\left(\text{1}\text{.0 }\mu \text{C}\times \frac{{10}^{-6}\text{ C}}{1\mu \text{C}}\right)\left(\text{2}\text{.5 }\mu \text{C}\times \frac{{10}^{-6}\text{ C}}{1\mu \text{C}}\right)}{{\left(0.10\text{ m}\right)}^2}\\ F_{\text{E}}=2.25\text{ N}\end{array}$

Conclusion:

Thus, the rank of the pairs of point charges according to the magnitude of electrostatic force is,

E. The force of a $1.0\mu \text{C}$ charge and a $2.5\mu \text{C}$ charge is $F_{\text{E}}=2.25\text{ N}$ .

A. The force of two $7.0\text{ nC}$ charges is $F_{\text{A}}=1.10\times {10}^{-5}\text{ N}$ .

B. The force of two $5.0\text{ nC}$ charges is $F_{\text{B}}=5.6\times {10}^{-6}\text{ N}$ .

C. The force of two $2.5\text{ nC}$ charges is $F_{\text{C}}=5.6\times {10}^{-6}\text{ N}$ .

D. The force of a $2.5\text{ nC}$ charge and a $5.0\text{ nC}$ charge is $F_{\text{D}}=2.8\times {10}^{-6}\text{ N}$ .

Thus, the ties are B and C.