A length-scale ($l$) depends on the permittivity ($\varepsilon$) of a dielectric material, Boltzmann constant ($k_{\text{B}}$), the absolute temperature ($T$), the number per unit volume ($n$) of certain charged particles, and the charge ($q$) carried by each of the particles. Which of the following expression(s) for $l$ is(are) dimensionally correct?

Step 1: Identify the dimensions of the given physical quantities.

The dimensions of the length-scale $l$, permittivity $\varepsilon$, Boltzmann constant $k_{\text{B}}$, absolute temperature $T$, number density $n$, and charge $q$ are determined from their physical definitions.

$$[l] = L$$ $$[\varepsilon] = \frac{[q^2]}{[F][r^2]} = M^{-1}L^{-3}T^4 I^2$$ $$[k_{\text{B}}] = \frac{[\text{energy}]}{[T]} = ML^2 T^{-2} \Theta^{-1}$$ $$[T] = \Theta$$ $$[n] = L^{-3}$$ $$[q] = IT$$

Step 2: Find the dimensions of key combinations.

Determine the dimension of the combination $\varepsilon k_{\text{B}} T$.

$$[\varepsilon k_{\text{B}} T] = (M^{-1}L^{-3}T^4 I^2)(ML^2 T^{-2} \Theta^{-1})(\Theta)$$ $$[\varepsilon k_{\text{B}} T] = L^{-1}T^2 I^2$$

Determine the dimension of the combination $\dfrac{q^2}{\varepsilon k_{\text{B}} T}$.

$$\left[\frac{q^2}{\varepsilon k_{\text{B}} T}\right] = \frac{I^2 T^2}{L^{-1}T^2 I^2}$$ $$\left[\frac{q^2}{\varepsilon k_{\text{B}} T}\right] = L$$

Step 3: Test the dimensional correctness of Option 1.

Check the dimension of the expression $\sqrt{\dfrac{nq^{2}}{\varepsilon k_{\text{B}} T}}$.

$$\left[\sqrt{\frac{nq^{2}}{\varepsilon k_{\text{B}} T}}\right] = \sqrt{[n] \cdot \left[\frac{q^2}{\varepsilon k_{\text{B}} T}\right]}$$ $$\left[\sqrt{\frac{nq^{2}}{\varepsilon k_{\text{B}} T}}\right] = \sqrt{L^{-3} \cdot L}$$ $$\left[\sqrt{\frac{nq^{2}}{\varepsilon k_{\text{B}} T}}\right] = L^{-1} \neq L$$

Step 4: Test the dimensional correctness of Option 2.

Check the dimension of the expression $\sqrt{\dfrac{\varepsilon k_{\text{B}} T}{nq^{2}}}$.

$$\left[\sqrt{\frac{\varepsilon k_{\text{B}} T}{nq^{2}}}\right] = \sqrt{\frac{1}{[n]} \cdot \left[\frac{\varepsilon k_{\text{B}} T}{q^2}\right]}$$ $$\left[\sqrt{\frac{\varepsilon k_{\text{B}} T}{nq^{2}}}\right] = \sqrt{L^3 \cdot L^{-1}}$$ $$\left[\sqrt{\frac{\varepsilon k_{\text{B}} T}{nq^{2}}}\right] = L$$

Step 5: Test the dimensional correctness of Option 3.

Check the dimension of the expression $\sqrt{\dfrac{q^{2}}{\varepsilon n^{2/3} k_{\text{B}} T}}$.

$$\left[\sqrt{\frac{q^{2}}{\varepsilon n^{2/3} k_{\text{B}} T}}\right] = \sqrt{[n^{-2/3}] \cdot \left[\frac{q^2}{\varepsilon k_{\text{B}} T}\right]}$$ $$\left[\sqrt{\frac{q^{2}}{\varepsilon n^{2/3} k_{\text{B}} T}}\right] = \sqrt{L^2 \cdot L}$$ $$\left[\sqrt{\frac{q^{2}}{\varepsilon n^{2/3} k_{\text{B}} T}}\right] = L^{3/2} \neq L$$

Step 6: Test the dimensional correctness of Option 4.

Check the dimension of the expression $\sqrt{\dfrac{q^{2}}{\varepsilon n^{1/3} k_{\text{B}} T}}$.

$$\left[\sqrt{\frac{q^{2}}{\varepsilon n^{1/3} k_{\text{B}} T}}\right] = \sqrt{[n^{-1/3}] \cdot \left[\frac{q^2}{\varepsilon k_{\text{B}} T}\right]}$$ $$\left[\sqrt{\frac{q^{2}}{\varepsilon n^{1/3} k_{\text{B}} T}}\right] = \sqrt{L \cdot L}$$ $$\left[\sqrt{\frac{q^{2}}{\varepsilon n^{1/3} k_{\text{B}} T}}\right] = L$$

Answer: The dimensionally correct expressions are (2) $\sqrt{\dfrac{\varepsilon k_{\text{B}} T}{nq^{2}}}$ and (4) $\sqrt{\dfrac{q^{2}}{\varepsilon n^{1/3} k_{\text{B}} T}}$.