(a) E =\frac{1}{4 \pi \epsilon_{0}} \int \frac{\rho \hat{ᴫ}}{ᴫ^{2}}\left(1+\frac{ᴫ}{\lambda}\right) e^{-ᴫ / \lambda} d \tau.
(b) Yes. The field of a point charge at the origin is radial and symmetric, so ∇×E = 0, and hence this is also true (by superposition) for any collection of charges.
(c) V=-\int_{\infty}^{r} E \cdot d l =-\frac{1}{4 \pi \epsilon_{0}} q \int_{\infty}^{r} \frac{1}{r^{2}}\left(1+\frac{r}{\lambda}\right) e^{-r / \lambda} d r
=\frac{1}{4 \pi \epsilon_{0}} q \int_{r}^{\infty} \frac{1}{r^{2}}\left(1+\frac{r}{\lambda}\right) e^{-r / \lambda} d r=\frac{q}{4 \pi \epsilon_{0}}\left\{\int_{r}^{\infty} \frac{1}{r^{2}} e^{-r / \lambda} d r+\frac{1}{\lambda} \int_{r}^{\infty} \frac{1}{r} e^{-r / \lambda} d r\right\} .
Now \int \frac{1}{r^{2}} e^{-r / \lambda} d r=-\frac{e^{-r / \lambda}}{r}-\frac{1}{\lambda} \int \frac{e^{-r / \lambda}}{r} d r \longleftarrow exactly right to kill the last term. Therefore
V(r)=\frac{q}{4 \pi \epsilon_{0}}\left\{-\left.\frac{e^{-r / \lambda}}{r}\right|_{r} ^{\infty}\right\}=\frac{q}{4 \pi \epsilon_{0}} \frac{e^{-r / \lambda}}{r} .
(d) \oint_{ S } E \cdot d a =\frac{1}{4\pi \varepsilon _0}q\frac{1}{R^2} \left(1+\frac{R}{\lambda}\right) e^{-R / \lambda}4\pi R^2=\frac{q}{\epsilon_{0}}\left(1+\frac{R}{\lambda}\right) e^{-R / \lambda} .
\int_{ \nu } V d \tau=\frac{q}{4\pi \varepsilon _0} \int_{0}^{R} \frac{e^{-r / \lambda}}{r} r^{2}4\pi d r=\frac{q}{\epsilon_{0}} \int_{0}^{R} r e^{-r / \lambda} d r=\frac{q}{\epsilon_{0}}\left[\frac{e^{-r / \lambda}}{(1 / \lambda)^{2}}\left(-\frac{r}{\lambda}-1\right)\right]_{0}^{R} .
=\lambda^{2} \frac{q}{\epsilon_{0}}\left\{-e^{-R / \lambda}\left(1+\frac{R}{\lambda}\right)+1\right\} .
\therefore \oint_{ S } E \cdot d a +\frac{1}{\lambda^{2}} \int_{ \nu } V d \tau=\frac{q}{\epsilon_{0}}\left\{\left(1+\frac{R}{\lambda}\right) e^{-R / \lambda}-\left(1+\frac{R}{\lambda}\right) e^{-R / \lambda}+1\right\}=\frac{q}{\epsilon_{0}} . qed
(e) Does the result in (d) hold for a nonspherical surface? Suppose we
make a “dent” in the sphere—pushing a patch \left(\text { area } R^{2} \sin \theta d \theta d \phi\right) from radius R out to radius S \left(\operatorname{area} S^{2} \sin \theta d \theta d \phi\right) .
\Delta \oint E \cdot d a =\frac{q}{4 \pi \epsilon_{0}}\left\{\frac{1}{S^{2}}\left(1+\frac{S}{\lambda}\right) e^{-S / \lambda}\left(S^{2} \sin \theta d \theta d \phi\right)-\frac{1}{R^{2}}\left(1+\frac{R}{\lambda}\right) e^{-R / \lambda}\left(R^{2} \sin \theta d \theta d \phi\right)\right\}
=\frac{q}{4 \pi \epsilon_{0}}\left[\left(1+\frac{S}{\lambda}\right) e^{-S / \lambda}-\left(1+\frac{R}{\lambda}\right) e^{-R / \lambda}\right] \sin \theta d \theta d \phi .
\Delta \frac{1}{\lambda^{2}} \int V d \tau=\frac{1}{\lambda^{2}} \frac{q}{4 \pi \epsilon_{0}} \int \frac{e^{-r / \lambda}}{r} r^{2} \sin \theta d r d \theta d \phi=\frac{1}{\lambda^{2}} \frac{q}{4 \pi \epsilon_{0}} \sin \theta d \theta d \phi \int_{R}^{S} r e^{-r / \lambda} d r
=-\left.\frac{q}{4 \pi \epsilon_{0}} \sin \theta d \theta d \phi\left(e^{-r / \lambda}\left(1+\frac{r}{\lambda}\right)\right)\right|_{R} ^{S}
=-\frac{q}{4 \pi \epsilon_{0}}\left[\left(1+\frac{S}{\lambda}\right) e^{-S / \lambda}-\left(1+\frac{R}{\lambda}\right) e^{-R / \lambda}\right] \sin \theta d \theta d \phi .
So the change in \frac{1}{\lambda^{2}} \int V d \tau exactly compensates for the change in \oint E \cdot d a , and we get \frac{1}{\epsilon_{0}} q for the total using the dented sphere, just as we did with the perfect sphere. Alny closed surface can be built up by successive distortions of the sphere, so the result holds for all shapes. By superposition, if there are many charges inside, the total is \frac{1}{\epsilon_{0}} Q_{\text {enc }} . Charges outside do not contribute (in the argument above we found that for this volume \oint E \cdot d a +\frac{1}{\lambda^{2}} \int V d \tau=0 __ and, again, the sum is not changed by distortions of the surface, as \text { long as } q \text { remains outside). So the new "Gauss's Law" holds for any charge configuration. }
(f) In di↵erential form, “Gauss’s law” reads: \nabla \cdot E +\frac{1}{\lambda^{2}} V=\frac{1}{\epsilon_{0}} \rho , or, putting it all in terms of E:
\nabla \cdot E -\frac{1}{\lambda^{2}} \int E \cdot d l =\frac{1}{\epsilon_{0}} \rho . \text { Since } E =- \nabla V , this also yields “Poisson’s equation”: -\nabla^{2} V+\frac{1}{\lambda^{2}} V=\frac{1}{\epsilon_{0}} \rho.
(g) Refer to ”Gauss’s law” in di↵erential form (f). Since E is zero, inside a conductor (otherwise charge would move, and in such a direction as to cancel the field), V is constant (inside), and hence ρ is uniform, throughout the volume. Any “extra” charge must reside on the surface. (The fraction at the surface depends on λ, and on the shape of the conductor.)
