Chapter 8, Problem 8.84E

Interpretation Introduction

Interpretation:

The varying values of force on two unit charges, between a vacuum and some medium with a nonzero dielectric constant, are to be calculated. The same evaluations obtained for charges of same sign are to be explained and the results are to be compared with charges of opposite sign.

Concept introduction:

The force of attraction between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of distance between them. The formula for determining force of attraction is given below.

$F=\frac{q_1{q}_2}{4{\text{πε}}_{\text{o}}{\text{ε}}_{\text{r}}{r}^2}$

In the above equation, $F$ is the force of attraction between two charges, $q_1$ and $q_2$ are the charges, $r$ is the distance between charges, ${\text{ε}}_{\text{r}}$ is the dielectric constant of the medium and ${\text{ε}}_{\text{o}}$ is the permittivity of free space. The value of ${\text{ε}}_{\text{o}}$ is defined as $8.854\times {10}^{-12}{\text{ C}}^{\text{2}}{\text{ J}}^{-\text{1}}{\text{m}}^{-\text{1}}$.

Answer to Problem 8.84E

The varying values of force on two unit charges between a vacuum and some medium with a nonzero dielectric constant are calculated. The same evaluations are obtained for charges of same sign but having positive values. The results with charges of opposite sign are compared with that of same sign.

Explanation of Solution

The expression of force between unit charges of opposite sign in vacuum can be written as follows.

$F=\frac{-1}{4\times 3.14\times 8.854\times {10}^{-12}\times r^2}$

The expression of force between unit charges of opposite sign in medium (taking water) having dielectric constant $\left(78\right)$ can be written as follows.

$F=\frac{-1}{4\times 3.14\times 8.854\times {10}^{-12}\times 78\times r^2}$

The table having the force between the two charges at distances ranging from $1\stackrel{\circ }{\mathrm{{\rm A}}}$ to $25\stackrel{\circ }{\mathrm{{\rm A}}}$ in $1\stackrel{\circ }{\mathrm{{\rm A}}}$ increments is given below.

Distance (m)	Force in vacuum(N)	Force in medium(N)
$1\times {10}^{-10}$	$-8.89\times {10}^{29}$	$-1.43\times {10}^{28}$
$2\times {10}^{-10}$	$-2.25\times {10}^{29}$	$-2.28\times {10}^{27}$
$3\times {10}^{-10}$	$-9.99\times {10}^{28}$	$-1.16\times {10}^{27}$
$4\times {10}^{-10}$	$-5.62\times {10}^{28}$	$-3.65\times {10}^{27}$
$5\times {10}^{-10}$	$-3.60\times {10}^{28}$	$-8.92\times {10}^{27}$
$6\times {10}^{-10}$	$-2.50\times {10}^{28}$	$-1.85\times {10}^{26}$
$7\times {10}^{-10}$	$-1.83\times {10}^{28}$	$-3.42\times {10}^{26}$
$8\times {10}^{-10}$	$-1.40\times {10}^{28}$	$-5.84\times {10}^{26}$
$9\times {10}^{-10}$	$-1.11\times {10}^{28}$	$-9.36\times {10}^{26}$
$10\times {10}^{-10}$	$-8.99\times {10}^{27}$	$-1.43\times {10}^{25}$
$11\times {10}^{-10}$	$-7.43\times {10}^{27}$	$-2.09\times {10}^{25}$
$12\times {10}^{-10}$	$-6.24\times {10}^{27}$	$-2.96\times {10}^{25}$
$13\times {10}^{-10}$	$-5.32\times {10}^{27}$	$-4.07\times {10}^{25}$
$14\times {10}^{-10}$	$-4.59\times {10}^{27}$	$-5.48\times {10}^{25}$
$15\times {10}^{-10}$	$-3.99\times {10}^{27}$	$-7.22\times {10}^{25}$
$16\times {10}^{-10}$	$-3.51\times {10}^{27}$	$-9.35\times {10}^{25}$
$17\times {10}^{-10}$	$-3.11\times {10}^{27}$	$-1.19\times {10}^{24}$
$18\times {10}^{-10}$	$-2.77\times {10}^{27}$	$-1.50\times {10}^{24}$
$19\times {10}^{-10}$	$-2.49\times {10}^{27}$	$-1.86\times {10}^{24}$
$20\times {10}^{-10}$	$-2.25\times {10}^{27}$	$-2.28\times {10}^{24}$
$21\times {10}^{-10}$	$-2.04\times {10}^{27}$	$-2.77\times {10}^{24}$
$22\times {10}^{-10}$	$-1.86\times {10}^{27}$	$-3.34\times {10}^{24}$
$23\times {10}^{-10}$	$-1.70\times {10}^{27}$	$-3.99\times {10}^{24}$
$24\times {10}^{-10}$	$-1.56\times {10}^{27}$	$-4.73\times {10}^{24}$
$25\times {10}^{-10}$	$-1.44\times {10}^{27}$	$-5.57\times {10}^{24}$

The expression of force between unit charges of same sign in vacuum can be written as follows.

$F=\frac{1}{4\times 3.14\times 8.854\times {10}^{-12}\times r^2}$

The expression of force between unit charges of same sign in medium (taking water) having dielectric constant $\left(78\right)$ can be written as follows.

$F=\frac{1}{4\times 3.14\times 8.854\times {10}^{-12}\times 78\times r^2}$

The table having the force between the two charges having same sign at distances ranging from $1\stackrel{\circ }{\mathrm{{\rm A}}}$ to $25\stackrel{\circ }{\mathrm{{\rm A}}}$ in $1\stackrel{\circ }{\mathrm{{\rm A}}}$ increments is given below.

Distance (m)	Force in vacuum(N)	Force in medium(N)
$1\times {10}^{-10}$	$8.89\times {10}^{29}$	$1.43\times {10}^{28}$
$2\times {10}^{-10}$	$2.25\times {10}^{29}$	$2.28\times {10}^{27}$
$3\times {10}^{-10}$	$9.99\times {10}^{28}$	$1.16\times {10}^{27}$
$4\times {10}^{-10}$	$5.62\times {10}^{28}$	$3.65\times {10}^{27}$
$5\times {10}^{-10}$	$3.60\times {10}^{28}$	$8.92\times {10}^{27}$
$6\times {10}^{-10}$	$2.50\times {10}^{28}$	$1.85\times {10}^{26}$
$7\times {10}^{-10}$	$1.83\times {10}^{28}$	$3.42\times {10}^{26}$
$8\times {10}^{-10}$	$1.40\times {10}^{28}$	$5.84\times {10}^{26}$
$9\times {10}^{-10}$	$1.11\times {10}^{28}$	$9.36\times {10}^{26}$
$10\times {10}^{-10}$	$8.99\times {10}^{27}$	$1.43\times {10}^{25}$
$11\times {10}^{-10}$	$7.43\times {10}^{27}$	$2.09\times {10}^{25}$
$12\times {10}^{-10}$	$6.24\times {10}^{27}$	$2.96\times {10}^{25}$
$13\times {10}^{-10}$	$5.32\times {10}^{27}$	$4.07\times {10}^{25}$
$14\times {10}^{-10}$	$4.59\times {10}^{27}$	$5.48\times {10}^{25}$
$15\times {10}^{-10}$	$3.99\times {10}^{27}$	$7.22\times {10}^{25}$
$16\times {10}^{-10}$	$3.51\times {10}^{27}$	$9.35\times {10}^{25}$
$17\times {10}^{-10}$	$3.11\times {10}^{27}$	$1.19\times {10}^{24}$
$18\times {10}^{-10}$	$2.77\times {10}^{27}$	$1.50\times {10}^{24}$
$19\times {10}^{-10}$	$2.49\times {10}^{27}$	$1.86\times {10}^{24}$
$20\times {10}^{-10}$	$2.25\times {10}^{27}$	$2.28\times {10}^{24}$
$21\times {10}^{-10}$	$2.04\times {10}^{27}$	$2.77\times {10}^{24}$
$22\times {10}^{-10}$	$1.86\times {10}^{27}$	$3.34\times {10}^{24}$
$23\times {10}^{-10}$	$1.70\times {10}^{27}$	$3.99\times {10}^{24}$
$24\times {10}^{-10}$	$1.56\times {10}^{27}$	$4.73\times {10}^{24}$
$25\times {10}^{-10}$	$1.44\times {10}^{27}$	$5.57\times {10}^{24}$

The forcebetween unit charges of same sign and that between opposite sign is different. Forcebetween unit charges of same sign represents repulsion while that between opposite sign represent attraction.

Conclusion

The varying values of force on two unit charges between a vacuum and some medium with a nonzero dielectric constant are calculated. The same evaluations are obtained for charges of same sign but having positive values due to difference in charge. The results with charges of opposite sign are compared with that of the values of same sign.