Imagine that new and extraordinarily precise measurements have revealed an error in Coulomb's law. The actual force of interaction between two point charges is found to be

$\mathbf{F}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}\left(1+\frac{r}{\lambda }\right){\mathbf{e}}^{\mathbf{-}\left(\frac{r}{\lambda }\right)}\widehat{\mathbf{r}}$

where λ is a new constant of nature (it has dimensions of length, obviously, and is a huge number—say half the radius of the known universe—so that the correction is small, which is why no one ever noticed the discrepancy before). You are charged with the task of reformulating electrostatics to accommodate the new discovery. Assume the principle of superposition still holds.

a. What is the electric field of a charge distribution ρ (replacing Eq. 2.8)?

b. Does this electric field admit a scalar potential? Explain briefly how you reached your conclusion. (No formal proof necessary—just a persuasive argument.)

c. Find the potential of a point charge q—the analog to Eq. 2.26. (If your answer to (b) was "no," better go back and change it!) Use ∞ as your reference point.

d. For a point charge q at the origin, show that

$\underset{\mathbf{S}}{\mathbf{\oint }}\mathbf{E}\mathbf{.}\mathbf{da}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{\lambda }}^{\mathbf{2}}}\underset{V}{\int }\mathrm{Vd\tau }\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\epsilon }}_{\mathbf{0}}}\mathbf{q}$

where S is the surface, V the volume, of any sphere centered at q.

e. Show that this result generalizes:

$\underset{\mathbf{S}}{\mathbf{\oint }}\mathbf{E}\mathbf{.}\mathbf{da}\mathbf{+}\frac{\mathbf{1}}{{\mathbf{\lambda }}^{\mathbf{2}}}\underset{V}{\int }\mathrm{Vd\tau }\mathbf{=}\frac{\mathbf{1}}{{\mathbf{\epsilon }}_{\mathbf{0}}}{\mathbf{Q}}_{\mathbf{enc}}$

for any charge distribution. (This is the next best thing to Gauss's Law, in the new "electrostatics.”)

f. Draw the triangle diagram (like Fig. 2.35) for this world, putting in all the appropriate formulas. (Think of Poisson's equation as the formula for ρ in terms of V, and Gauss's law (differential form) as an equation for ρ in terms of E.)

g. Show that some of the charge on a conductor distributes itself (uniformly!) over the volume, with the remainder on the surface. [Hint: E is still zero, inside a conductor.]

Coulomb's law is the most fundamental law’s in electrostatics, which mainly deals with force of attraction and repulsion between charges kept in the vicinity. The force is inversely related to the distance between the charges and directly proportional to the product of the magnitude of the charges kept in the vicinity.

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

(a)

Consider the equation for the force as:

$\mathrm{F}=\mathrm{qE}$ ……. (1)

Here, F is the force, E is the electric field intensity,

As given in the question:

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}\left(1+\frac{\mathrm{r}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{r}}{\mathrm{\lambda }}\right)}\widehat{\mathrm{r}}$

Therefore, from equation (i), we can write

$\begin{array}{c}\mathrm{E}=\frac{\mathrm{F}}{\mathrm{q}}\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\int \frac{\mathrm{\rho }}{{\mathrm{r}}^2}\left(1+\frac{\mathrm{r}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{r}}{\mathrm{\lambda }}\right)}\widehat{\mathrm{r}}\mathrm{d\tau }\end{array}$

(b)

Yes, the electric field admits scaler potential because, when a point charge is placed at the origin of any coordinate system, its field will always be in a radial direction; hence,$\nabla \times \mathrm{E}=0$ , which is also true for the collection of the charges.

(c)

Consider the formulae for calculating potential as follows:

$\mathrm{V}=-\underset{\mathrm{\infty }}{\overset{\mathrm{r}}{\int }}\mathrm{E}.\mathrm{dl}$

Substitute value of E and solve as:

$\begin{array}{c}\mathrm{V}=-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\mathrm{q}\underset{\mathrm{\infty }}{\overset{\mathrm{r}}{\int }}\frac{1}{{\mathrm{r}}^2}\left(1+\frac{\mathrm{r}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{r}}{\mathrm{\lambda }}\right)}\mathrm{dr}\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\mathrm{q}\underset{\mathrm{\infty }}{\overset{\mathrm{r}}{\int }}\frac{1}{{\mathrm{r}}^2}\left(1+\frac{\mathrm{r}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{r}}{\mathrm{\lambda }}\right)}\mathrm{dr}\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\mathrm{q}\left\{\underset{\mathrm{r}}{\overset{\mathrm{\infty }}{\int }}\frac{1}{{\mathrm{r}}^2}{\mathrm{e}}^{-\frac{\mathrm{r}}{\mathrm{\lambda }}}\mathrm{dr}+\frac{1}{\mathrm{\lambda }}\underset{\mathrm{r}}{\overset{\mathrm{\infty }}{\int }}\frac{1}{\mathrm{r}}{\mathrm{e}}^{-\frac{\mathrm{r}}{\mathrm{\lambda }}}\mathrm{dr}\right\}\end{array}$

Therefore, the potential of the point charge is:

$\mathrm{V}\left(\mathrm{r}\right)=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left\{{\overline{)-\frac{1}{\mathrm{r}}{\mathrm{e}}^{-\frac{\mathrm{r}}{\mathrm{\lambda }}}}}_{\mathrm{r}}^{\mathrm{\infty }}\right\}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{e}}^{-\frac{\mathrm{r}}{\mathrm{\lambda }}}}{\mathrm{r}}$

(d)

Solve for the integral as:

Consider to determine the volume integral of the potential (V) solve as:

$\begin{array}{c}\underset{\mathrm{V}}{\int }\mathrm{Vd\tau }=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\underset{0}{\overset{\mathrm{R}}{\int }}\frac{{\mathrm{e}}^{-\frac{\mathrm{r}}{\mathrm{\lambda }}}}{\mathrm{r}}{\mathrm{r}}^2\mathrm{dr}\\ =\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}{\left[\frac{{\mathrm{e}}^{-\frac{\mathrm{r}}{\mathrm{\lambda }}}}{{(1/\mathrm{\lambda })}^2}\left(-\frac{\mathrm{r}}{\mathrm{\lambda }}-1\right)\right]}_0^{\mathrm{R}}\\ ={\mathrm{\lambda }}^2\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}\left\{-{\mathrm{e}}^{-\frac{\mathrm{R}}{\mathrm{\lambda }}}\left(\frac{\mathrm{R}}{\mathrm{\lambda }}+1\right)+1\right\}\end{array}$

Therefore, the value of the

$\begin{array}{c}\underset{\mathrm{S}}{\oint }\mathrm{E}.\mathrm{da}+\frac{1}{{\mathrm{\lambda }}^2}\underset{\mathrm{V}}{\int }\mathrm{Vd\tau }=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}\left(1+\frac{\mathrm{R}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{R}}{\mathrm{\lambda }}\right)}+\frac{1}{{\mathrm{\lambda }}^2}{\mathrm{\lambda }}^2\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}\left\{-{\mathrm{e}}^{-\frac{\mathrm{R}}{\mathrm{\lambda }}}\left(\frac{\mathrm{R}}{\mathrm{\lambda }}+1\right)+1\right\}\\ =\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}\left\{\left(1+\frac{\mathrm{R}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{R}}{\mathrm{\lambda }}\right)}-\left(1+\frac{\mathrm{R}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{R}}{\mathrm{\lambda }}\right)}+1\right\}\\ =\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}\end{array}$

(e)

Consider in spherical coordinate system with a Radius (S), the radius is changed to (R) due to distortion in the field. Therefore, we can write:

$\begin{array}{c}\mathrm{\Delta }\oint \mathrm{E}.\mathrm{da}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left\{\frac{1}{{\mathrm{S}}^2}\left(1+\frac{\mathrm{S}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{S}}{\mathrm{\lambda }}\right)}\left({\mathrm{S}}^2\mathrm{sin\theta d\theta d\varphi }\right)-\frac{1}{{\mathrm{R}}^2}\left(1+\frac{\mathrm{R}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{R}}{\mathrm{\lambda }}\right)}\left({\mathrm{R}}^2\mathrm{sin\theta d\theta d\varphi }\right)\right\}\\ =\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left[\left(1+\frac{\mathrm{S}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{S}}{\mathrm{\lambda }}\right)}-\left(1+\frac{\mathrm{R}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{R}}{\mathrm{\lambda }}\right)}\right]\mathrm{sin\theta d\theta d\varphi }\end{array}$

Solve for the change in potential as:

$\begin{array}{c}\mathrm{\Delta }\frac{1}{{\mathrm{\lambda }}^2}\underset{\mathrm{V}}{\int }\mathrm{Vd\tau }=\frac{1}{{\mathrm{\lambda }}^2}\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\int \frac{{\mathrm{e}}^{-\frac{\mathrm{r}}{\mathrm{\lambda }}}}{\mathrm{r}}{\mathrm{r}}^2\mathrm{sin\theta drd\theta d\varphi }\\ =\frac{1}{{\mathrm{\lambda }}^2}\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\mathrm{sin\theta d\theta d\varphi }\underset{\mathrm{R}}{\overset{\mathrm{S}}{\int }}{\mathrm{re}}^{-\frac{\mathrm{r}}{\mathrm{\lambda }}}\mathrm{dr}\\ =-\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\mathrm{sin\theta d\theta d\varphi }{\overline{)\left({\mathrm{e}}^{-\frac{\mathrm{r}}{\mathrm{\lambda }}}\left(1+\frac{\mathrm{r}}{\mathrm{\lambda }}\right)\right)}}_{\mathrm{R}}^{\mathrm{S}}\\ =-\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0}\left[\left(1+\frac{\mathrm{S}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{S}}{\mathrm{\lambda }}\right)}-\left(1+\frac{\mathrm{R}}{\mathrm{\lambda }}\right){\mathrm{e}}^{-\left(\frac{\mathrm{R}}{\mathrm{\lambda }}\right)}\right]\mathrm{sin\theta d\theta d\varphi }\end{array}$

Therefore, solve for the potential as:

$\mathrm{\Delta }\oint \mathrm{E}.\mathrm{da}+\mathrm{\Delta }\frac{1}{{\mathrm{\lambda }}^2}\underset{\mathrm{V}}{\int }\mathrm{Vd\tau }=0$

This means by distorting the field there is no change in sum has been observed. Therefore, this result is generalized for any charge distribution.

(f)

The triangle diagram is shown below:

Consider the equation for the potential from the Gauss law as:

$\nabla \mathrm{E}+\frac{1}{{\mathrm{\lambda }}^2}\mathrm{V}=\frac{\mathrm{\rho }}{{\mathrm{\epsilon }}_0}$

Rewrite the equation as:

$\nabla \mathrm{E}+\frac{1}{{\mathrm{\lambda }}^2}\int \mathrm{E}.\mathrm{dl}=\frac{\mathrm{\rho }}{{\mathrm{\epsilon }}_0}$

Since $\mathrm{E}=-\nabla \mathrm{V}$.

Therefore,

$\nabla \mathrm{E}=-{\nabla }^2\mathrm{V}$

Hence,

$-{\nabla }^2\mathrm{V}+\frac{1}{{\mathrm{\lambda }}^2}\mathrm{V}=\frac{\mathrm{\rho }}{{\mathrm{\epsilon }}_0}$

(g)

Consider that for a conductor, Electric field intensity is always zero inside the conductor, and the electric field is only present outside the conductor. The potential (V) is the same inside and outside the conductor; hence, charge density is uniform throughout the conductor's volume, so if we put any extra charge by external means, it will always be present conductor's surface.