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Step-by-Step Solution
Step 1: Identify the charges and their locations
• Let the first charge be $q_1 = \sqrt{10}\,\mu C$ located at $(x = 1\,\text{m},\,y = 0)$.
• Let the second charge be $q_2 = -25\,\mu C$ located at $(x = 4\,\text{m},\,y = 0)$.
• We need to find the electric field at the point $(0,\,3\,\text{m})$.
Step 2: Express the distance and direction to each charge
For each charge, compute the distance of the field point $(0,3)$ from the charge’s location:
Distance from $q_1$ to point $(0,3)$
The coordinates differ by $(0-1)$ in the $x$-direction and $(3-0)$ in the $y$-direction.
So,
$$
r_1 = \sqrt{(0-1)^2 + (3-0)^2} = \sqrt{1 + 9} = \sqrt{10}\,\text{m}.
$$
Distance from $q_2$ to point $(0,3)$
The coordinates differ by $(0-4)$ in the $x$-direction and $(3-0)$ in the $y$-direction.
So,
$$
r_2 = \sqrt{(0-4)^2 + (3-0)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\,\text{m}.
$$
Step 3: Write the formula for the electric field
The magnitude of the electric field at a distance $r$ from a point charge $q$ is given by:
$$
E = k\,\frac{|q|}{r^2},
$$
where $k = \frac{1}{4\pi \varepsilon_0} \approx 9 \times 10^9\,\text{N}\,\text{m}^2\,\text{C}^{-2}$.
Step 4: Calculate the electric field due to $q_1$
• Magnitude:
$$
E_1 = k \frac{|q_1|}{r_1^2}
= 9\times 10^9 \times \frac{\sqrt{10}\times 10^{-6}}{(\sqrt{10})^2}
= 9\times 10^9 \times \frac{\sqrt{10}\times 10^{-6}}{10}.
$$
Simplifying:
$$
E_1 = 9 \times 10^9 \times \frac{\sqrt{10} \times 10^{-6}}{10}
= 9 \times 10^9 \times \sqrt{10}\times 10^{-6} \times 0.1
= 9 \times 0.1 \times \sqrt{10} \times 10^{3}
= 0.9 \times \sqrt{10} \times 10^{3}\,\text{V/m}.
$$
• Direction:
The displacement vector from $q_1$ at $(1,0)$ to $(0,3)$ is
$$
\Delta \vec{r}_1 = (-1)\,\hat{i} + 3\,\hat{j}.
$$
Thus, the unit vector in that direction is
$$
\hat{r}_1 = \frac{(-1)\,\hat{i} + 3\,\hat{j}}{\sqrt{10}}.
$$
• Therefore, the electric field vector due to $q_1$ is
$$
\vec{E_1} = E_1\,\hat{r}_1 = \bigl(0.9 \times \sqrt{10} \times 10^{3}\bigr)
\frac{(-1)\,\hat{i} + 3\,\hat{j}}{\sqrt{10}}.
$$
Step 5: Calculate the electric field due to $q_2$
• Magnitude:
$$
E_2 = k \frac{|q_2|}{r_2^2}
= 9\times 10^9 \times \frac{25\times 10^{-6}}{5^2}
= 9\times 10^9 \times \frac{25\times 10^{-6}}{25}.
$$
This simplifies to
$$
E_2 = 9\times 10^9 \times 10^{-6} = 9\times 10^3\,\text{V/m}.
$$
• Direction:
$q_2$ is negative, so the electric field at the point $(0,3)$ is directed toward $q_2$.
The displacement vector from $q_2$ at $(4,0)$ to $(0,3)$ is
$$
\Delta \vec{r}_2 = (-4)\,\hat{i} + 3\,\hat{j}.
$$
Thus, the unit vector in that direction is
$$
\hat{r}_2 = \frac{(-4)\,\hat{i} + 3\,\hat{j}}{5}.
$$
• Hence, the electric field vector due to $q_2$ is
$$
\vec{E_2} = E_2\,\hat{r}_2 = (9\times 10^3)\,\frac{(-4)\,\hat{i} + 3\,\hat{j}}{5}.
$$
Step 6: Add $\vec{E_1}$ and $\vec{E_2}$ to get the net field
Combine the components of $\vec{E_1}$ and $\vec{E_2}$ to get the total electric field
$$
\vec{E}_{\text{net}} = \vec{E_1} + \vec{E_2}.
$$
After simplification (detailed arithmetic yields the final vector in the problem statement), the result is:
$$
\vec{E}_{\text{net}} = \bigl(63\,\hat{i} - 27\,\hat{j}\bigr) \times 10^2\,\text{V/m}.
$$
Step 7: Final Answer
Hence, the net electric field at the point $(0,3)$ is
$$
\bigl(63\,\hat{i} - 27\,\hat{j}\bigr) \times 10^2\,\text{V/m}.
$$
