1. **Problem statement:** We are given a uniform surface charge density $\sigma = 6$ kC/m$^2$ on a semi-infinite metal strip and asked to find the electric field density at a two-and-a-half right angle corner field, with the dielectric constant $\varepsilon = 2$. The answers are given in terms of $\varepsilon_0$ in units of $\Omega$/m$^3$. 2. **Understanding the problem:** The problem involves calculating the electric field density near a corner with a given surface charge density and dielectric constant. The dielectric constant modifies the permittivity as $\varepsilon_r = 2$, so the effective permittivity is $\varepsilon = \varepsilon_r \varepsilon_0 = 2 \varepsilon_0$. 3. **Formula and rules:** The electric field $E$ due to a surface charge density $\sigma$ in a medium with permittivity $\varepsilon$ is generally given by: $$E = \frac{\sigma}{\varepsilon}$$ However, the problem involves a corner with an angle of $2.5 \times 90^\circ = 225^\circ$, which affects the field distribution. The field near a corner with angle $\alpha$ in a dielectric medium is proportional to $\frac{\pi}{\alpha \varepsilon}$ times the surface charge density. 4. **Calculate the angle in radians:** $$\alpha = 2.5 \times 90^\circ = 225^\circ = \frac{225 \pi}{180} = \frac{5\pi}{4}$$ 5. **Calculate the electric field density:** $$E = \frac{\pi}{\alpha \varepsilon} \sigma = \frac{\pi}{\frac{5\pi}{4} \times 2 \varepsilon_0} \times 6 = \frac{\pi}{\frac{5\pi}{4} \times 2 \varepsilon_0} \times 6$$ Simplify denominator: $$\frac{5\pi}{4} \times 2 = \frac{5\pi}{4} \times \frac{8}{4} = \frac{10\pi}{4} = \frac{5\pi}{2}$$ Actually, better to multiply directly: $$\frac{5\pi}{4} \times 2 = \frac{5\pi}{4} \times \frac{2}{1} = \frac{10\pi}{4} = \frac{5\pi}{2}$$ So: $$E = \frac{\pi}{\frac{5\pi}{2} \varepsilon_0} \times 6 = \frac{\pi}{\frac{5\pi}{2} \varepsilon_0} \times 6 = \frac{6}{\frac{5}{2} \varepsilon_0} = \frac{6 \times 2}{5 \varepsilon_0} = \frac{12}{5 \varepsilon_0}$$ 6. **Express as a fraction with denominator $\varepsilon_0$:** $$E = \frac{12}{5 \varepsilon_0} = \frac{12/5}{\varepsilon_0} = \frac{2.4}{\varepsilon_0}$$ 7. **Compare with given options:** Options are: 1) $\frac{3}{\varepsilon_0}$ 2) $\frac{9}{4 \varepsilon_0} = 2.25/\varepsilon_0$ 3) $\frac{16}{\varepsilon_0} = 16/\varepsilon_0$ 4) $\frac{4}{\varepsilon_0} = 4/\varepsilon_0$ Our calculated value $2.4/\varepsilon_0$ is closest to option 2) $2.25/\varepsilon_0$. **Final answer:** Option 2) $\frac{9}{4 \varepsilon_0}$