1. **Problem statement:** Two point charges each of $2.19\,\mu C$ are located on the x-axis at $x=1.13\,m$ and $x=-1.13\,m$. We need to find the electric field at the point on the y-axis at $y=0.474\,m$. 2. **Formula for electric field due to a point charge:** The electric field $\vec{E}$ at a distance $r$ from a point charge $q$ is given by $$\vec{E} = \frac{k |q|}{r^2} \hat{r}$$ where $k = 8.99 \times 10^9 \ N \cdot m^2/C^2$ is Coulomb's constant, and $\hat{r}$ is the unit vector from the charge to the point of interest. 3. **Calculate distance from each charge to the point on y-axis:** Each charge is at $(\pm 1.13, 0)$ and the point is at $(0, 0.474)$. Distance $r = \sqrt{(1.13)^2 + (0.474)^2} = \sqrt{1.2769 + 0.2245} = \sqrt{1.5014} \approx 1.225\,m$. 4. **Calculate magnitude of electric field from one charge:** $$E = \frac{8.99 \times 10^9 \times 2.19 \times 10^{-6}}{(1.225)^2} = \frac{1.968 \times 10^4}{1.5014} \approx 1.31 \times 10^4 \ N/C$$ 5. **Determine direction of electric field vectors:** The two charges are positive, so the electric field at the point due to each charge points away from the charge. The x-components of the fields from the two charges cancel because they are equal in magnitude and opposite in direction. The y-components add up. 6. **Calculate y-component of electric field from one charge:** The angle $\theta$ between the line connecting charge to point and the x-axis is $$\theta = \tan^{-1}\left(\frac{0.474}{1.13}\right) \approx 22.8^\circ$$ 7. **Calculate total electric field on y-axis:** The y-component from one charge is $$E_y = E \sin \theta = 1.31 \times 10^4 \times \sin 22.8^\circ \approx 1.31 \times 10^4 \times 0.388 = 5.08 \times 10^3 \ N/C$$ Since there are two charges, total $E_y = 2 \times 5.08 \times 10^3 = 1.02 \times 10^4 \ N/C$. **Final answer:** The electric field at $y=0.474\,m$ on the y-axis is approximately $$\boxed{1.02 \times 10^4 \ N/C \text{ directed along positive y-axis}}$$