Solution
The force is given by:
$$\vec{F} = \frac{Qq(1 – \sqrt{\alpha r})}{4\pi \epsilon_0 r^3} \vec{r} = \frac{Qq(1 – \sqrt{\alpha r})}{4\pi \epsilon_0 r^2} \hat{r}$$Analysis of Options:
(a) Electric Field: Since $\vec{F} = q\vec{E}$, the electric field due to Q is:
$$\vec{E} = \frac{\vec{F}}{q} = \frac{Q(1 – \sqrt{\alpha r})}{4\pi \epsilon_0 r^3} \vec{r}$$This statement is Correct.
(b) Line Integral / Conservative Nature: The force is a central force because it acts along the radial direction $\hat{r}$ and its magnitude depends only on $r$. All central forces are conservative. Therefore, the line integral over a closed path is zero ($\oint \vec{E} \cdot d\vec{l} = 0$).
This statement is Correct.
(c) Gauss’s Law: Gauss’s Law ($\oint \vec{E} \cdot d\vec{A} = Q/\epsilon_0$) implies that the flux is independent of the radius of the Gaussian surface. This requires $E \propto 1/r^2$. Here, the field has an additional term dependent on $r^{-1.5}$. The flux will vary with radius. Thus, the standard Gauss’s law does not hold in free space for this field.
This statement is Incorrect.
(d) Non-conservative: As established in (b), the field is conservative.
This statement is Incorrect.
Answer: (a) and (b)