Chapter 3.1, Problem 3.1.3SR

Arrange the following pairs of charged particles in order of increasing magnitude of electrostatic attraction (E_el).

(a) i < ii < iii < iv

(b) iv < iii < ii < i

(c) i = iii < ii < iv

(d) ii < i = iii < iv

(e) iv < i = ii < iii

Interpretation Introduction

Interpretation:

The electrostatic potential energies between the given charges in the given statement should be compared by using the concept of electrostatic energy.

Concept Introduction:

Energy is the ability to do work or transfer heat where work is the movement of a body using some force.  The SI unit of energy is joule ($\text{J}$).  Energy can be in the form of kinetic energy or potential energy.  Potential energy is the energy possessed by virtue of position. Other forms of potential energy are chemical energy and electrostatic energy.

Electrostatic energy is potential energy which results from the interaction of charged particles.  Oppositely charged particles attract each other and particles of like charges repel each other.  The magnitude of the resulting electrostatic potential energy is proportional to the product of the two charges (${\text{Q}}_{\text{1}}{\text{ and Q}}_{\text{2}}$) divided by the distance between the two charges ($\text{d}$).

${\text{E}}_{\text{el}}\propto \frac{{\text{Q}}_{\text{1}}{\text{Q}}_{\text{2}}}{\text{d}}$

If the charges ${\text{Q}}_{\text{1}}{\text{ and Q}}_{\text{2}}$ are opposite (i.e., one positive and one negative), ${\text{E}}_{\text{el}}$ gets a negative value which indicates attraction.  Like charges (i.e., either both positive or both negative) result in a positive value for ${\text{E}}_{\text{el}}$ which indicates repulsion.

To calculate: Determine the ${\text{E}}_{\text{el}}$ values between the given charges and compare them

Answer to Problem 3.1.3SR

The given pairs of charged particles in order of increasing magnitude of electrostatic attraction (${\text{E}}_{\text{el}}$) are arranged as $\text{i }=\text{ iii < ii < iv}$ which is the option (c).

Explanation of Solution

Reason for correct option

The electrostatic energy between two charges is calculated using the formulae: ${\text{E}}_{\text{el}}\propto \frac{{\text{Q}}_{\text{1}}{\text{Q}}_{\text{2}}}{\text{d}}$

It is used to compare the magnitudes of the ${\text{E}}_{\text{el}}$ values.  The ${\text{E}}_{\text{el}}$ values are easily calculated without knowing the distance.  Both the distance and the proportionality constant cancel in the formulae.

For charges +1 and −3:

$\begin{array}{l}{\text{E}}_{\text{el(+1,}-\text{3)}}=\text{ c}\frac{{\text{Q}}_{\text{1}}{\text{Q}}_{\text{2}}}{\text{d}}{\text{,Q}}_{\text{1}}{\text{ = +1 and Q}}_{\text{2}}\text{ = }-\text{3}\\ {\text{E}}_{\text{el(+1,}-\text{3)}}=\text{ c}\frac{\text{1 × }\left(-\text{3}\right)}{\text{d}}\\ \text{Skip the constants the negative sign, c and d,}\\ {\text{E}}_{\text{el(+1,}-\text{3)}}=\text{ 3}\end{array}$

For charges +2 and −2:

$\begin{array}{l}{\text{E}}_{\text{el(+2,}-\text{2)}}=\text{ c}\frac{{\text{Q}}_{\text{1}}{\text{Q}}_{\text{2}}}{\text{d}}{\text{, Q}}_{\text{1}}{\text{ = +2 and Q}}_{\text{2}}\text{ = }-\text{2}\\ {\text{E}}_{\text{el(+2,}-\text{2)}}=\text{ c}\frac{\text{2 × }\left(-\text{2}\right)}{\text{d}}\\ \text{Skip the constants the negative sign, c and d,}\\ {\text{E}}_{\text{el(+2,}-\text{2)}}=\text{ 4}\end{array}$

For charges +2 and −3:

$\begin{array}{l}{\text{E}}_{\text{el(+2,}-\text{3)}}=\text{ c}\frac{{\text{Q}}_{\text{1}}{\text{Q}}_{\text{2}}}{\text{d}}{\text{,Q}}_{\text{1}}{\text{ = +2 and Q}}_{\text{2}}\text{ = }-\text{3}\\ {\text{E}}_{\text{el(+2,}-\text{3)}}=\text{ c}\frac{\text{2 × }\left(-\text{3}\right)}{\text{2d}}\\ \text{Skip the constants the negative sign, c and d,}\\ {\text{E}}_{\text{el(+2,}-\text{3)}}=\text{ 3}\end{array}$

For charges +4 and −4:

$\begin{array}{l}{\text{E}}_{\text{el(+4,}-\text{4)}}=\text{ c}\frac{{\text{Q}}_{\text{1}}{\text{Q}}_{\text{2}}}{\text{d}}{\text{, Q}}_{\text{1}}{\text{ = +4 and Q}}_{\text{2}}\text{ = }-\text{4}\\ {\text{E}}_{\text{el(+4,}-\text{4)}}=\text{ c}\frac{\text{4 × }\left(-\text{4}\right)}{\text{3d}}\\ \text{Skip the constants the negative sign, c and d,}\\ {\text{E}}_{\text{el(+4,}-\text{4)}}=\text{ 5}\text{.3}\end{array}$

Therefore, the given pairs of charged particles in order of increasing magnitude of electrostatic attraction (${\text{E}}_{\text{el}}$) are arranged as $\text{i }=\text{ iii < ii < iv }\left(\text{3 }=\text{ 3 < 4 < 5}\text{.3}\right)$ which is the option (c).

Reasons for incorrect options:

By doing the calculations in which the given values are substituted, it is found that the option (a) is correct.  Therefore, the options (b), (c), (d) and (e) are incorrect.

Conclusion

The electrostatic potential energies between the given charges in the given statement are compared by using the concept of electrostatic energy.