Question

In the diagram below, there are two charges of +q and -q and six points (a through f) at various distances from the two charges. (Intro 1 figure) You will be asked to rank changes in the electric potential along paths between pairs of points.

Q: Using the diagram to the left, rank each of the given pathson the basis of the change in electric potential. Rank the largest-magnitude positive change (increase in electric potential)as largest and the largest-magnitude negative change (decrease in electric potential) as smallest.

Rank from largest to smallest. To rank items as equivalent, overlap them.

a. from c to b

b. from c to d

c. from c to e

d. from d to a

e. from b to a

f. from f to e

 

Answer 1

Concepts and reason

The concept required to solve the given question is work done in bringing a point charge from infinity, that is, electric potential.

The amount of work done needed to move a point charge from infinity to the specific point inside the field without producing any acceleration is termed as electric potential.

Initially, find out the electric potential at six different points and then rank the from largest to smallest.

Fundamentals

The expression of the electric potential from a point charge q at a distance r from the charge is given as follows:

$V = \frac{{kq}}{r}$

Here, k is the Coulomb’s constant, q is the charge, and r is the distance.

The expression of electric potential from one point (a) to another point (b) is given as follows:

${V_{ab}} = {V_b} - {V_a}$

Here, ${V_b}$is the electric potential at point b and ${V_a}$ is the electric potential at point a.

Since, the given charge distribution represents a dipole and point c and d is at the equatorial position of dipole. The electric potential at the equatorial position of dipole is zero. Therefore, potential at point c and d is zero.

${V_c} = {V_d} = 0$

The electric potential at point a due to two charges is given as follows:

$\begin{array}{c}\\{V_a} = \frac{{kq}}{{2r}} - \frac{{kq}}{{6r}}\\\\ = 0.33\frac{{kq}}{r}\\\end{array}$

The electric potential at point b due to two charges is given as follows:

$\begin{array}{c}\\{V_b} = \frac{{kq}}{r} - \frac{{kq}}{{5r}}\\\\ = 0.8\frac{{kq}}{r}\\\end{array}$

The electric potential at point e due to two charges is given as follows:

$\begin{array}{c}\\{V_e} = \frac{{kq}}{{5r}} - \frac{{kq}}{r}\\\\ = - 0.8\frac{{kq}}{r}\\\end{array}$

The electric potential at point f due to two charges is given as follows:

$\begin{array}{c}\\{V_f} = \frac{{kq}}{{6r}} - \frac{{kq}}{{2r}}\\\\ = - 0.33\frac{{kq}}{r}\\\end{array}$

The electric potential from c to b is given as follows:

${V_{cb}} = {V_b} - {V_c}$

Substitute $0.8\frac{{kq}}{r}$for ${V_b}$ and 0 for ${V_c}$ in the expression ${V_{cb}} = {V_b} - {V_c}$.

$\begin{array}{c}\\{V_{cb}} = 0.8\frac{{kq}}{r} - 0\\\\ = 0.8\frac{{kq}}{r}\\\end{array}$

The electric potential from c to d is given as follows:

${V_{cd}} = {V_d} - {V_c}$

Substitute 0 for ${V_d}$ and 0 for ${V_c}$ in the expression ${V_{cd}} = {V_d} - {V_c}$.

$\begin{array}{c}\\{V_{cd}} = 0 - 0\\\\ = 0\\\end{array}$

The electric potential from c to e is given as follows:

${V_{ce}} = {V_e} - {V_c}$

Substitute $- 0.8\frac{{kq}}{r}$ for ${V_e}$ and 0 for ${V_c}$ in the expression ${V_{ce}} = {V_e} - {V_c}$.

$\begin{array}{c}\\{V_{ce}} = - 0.8\frac{{kq}}{r} - 0\\\\ = - 0.8\frac{{kq}}{r}\\\end{array}$

The electric potential from d to a is given as follows:

${V_{da}} = {V_a} - {V_d}$

Substitute $0.333\frac{{kq}}{r}$ for ${V_a}$ and 0 for ${V_d}$ in the expression ${V_{da}} = {V_a} - {V_d}$.

$\begin{array}{c}\\{V_{da}} = 0.333\frac{{kq}}{r} - 0\\\\ = 0.333\frac{{kq}}{r}\\\end{array}$

The electric potential from b to a is given as follows:

${V_{ba}} = {V_a} - {V_b}$

Substitute $0.333\frac{{kq}}{r}$ for ${V_a}$ and $0.8\frac{{kq}}{r}$ for ${V_b}$ in the expression ${V_{ba}} = {V_a} - {V_b}$.

$\begin{array}{c}\\{V_{ba}} = 0.333\frac{{kq}}{r} - 0.8\frac{{kq}}{r}\\\\ = - 0.467\frac{{kq}}{r}\\\end{array}$

The electric potential from f to e is given as follows:

${V_{fe}} = {V_e} - {V_f}$

Substitute $- 0.8\frac{{kq}}{r}$ for ${V_e}$ and $- 0.333\frac{{kq}}{r}$ for ${V_f}$ in the expression ${V_{fe}} = {V_e} - {V_f}$.

$\begin{array}{c}\\{V_{fe}} = - 0.8\frac{{kq}}{r} - \left( { - 0.333\frac{{kq}}{r}} \right)\\\\ = - 0.467\frac{{kq}}{r}\\\end{array}$

Therefore, the electric potential is given as follows:

$\begin{array}{c}\\{V_{cb}} = 0.8\frac{{kq}}{r}\\\\{V_{cd}} = 0\\\\{V_{ce}} = - 0.8\frac{{kq}}{r}\\\end{array}$

$\begin{array}{c}\\{V_{da}} = 0.333\frac{{kq}}{r}\\\\{V_{ba}} = - 0.467\frac{{kq}}{r}\\\\{V_{fe}} = - 0.467\frac{{kq}}{r}\\\end{array}$

On comparing, the rank of electric potential is given as follows:

${V_{cb}} > {V_{da}} > {V_{cd}} > \left( {{V_{ba}} = {V_{fe}}} \right) > {V_{ce}}$

Ans:

The rank of electric potential from largest to smallest is ${V_{cb}} > {V_{da}} > {V_{cd}} > \left( {{V_{ba}} = {V_{fe}}} \right) > {V_{ce}}$.