1. Given data

• Charge $q_1 = \sqrt{10}\,\mu C = \sqrt{10}\times 10^{-6}\,C$
• Charge $q_2 = -25\,\mu C = -25\times 10^{-6}\,C$
• Position of $q_1: (1,0)$
• Position of $q_2: (4,0)$
• Field point $P: (0,3)$
• Coulomb constant: $k=\frac{1}{4\pi\varepsilon_0}=9\times 10^9\,\text{N m}^2\text{/C}^2$

We need the net electric field at $P$.

2. Electric field due to $q_1$

Vector from $q_1$ to $P$: $\vec r_1 = (0-1)\hat i + (3-0)\hat j = -\hat i + 3\hat j$

Magnitude: $r_1 = \sqrt{(-1)^2+3^2}=\sqrt{10}$

Electric field due to a point charge: $\vec E_1 = k\frac{q_1}{r_1^3}\vec r_1$

Substitute values: $\vec E_1 = 9\times 10^9 \cdot \frac{\sqrt{10}\times 10^{-6}}{(\sqrt{10})^3}(-\hat i+3\hat j)$

Since $\frac{\sqrt{10}}{(\sqrt{10})^3}=\frac{1}{10}$

So, $\vec E_1 = 9\times 10^9\times 10^{-6}\times \frac{1}{10}(-\hat i+3\hat j)$ $\vec E_1 = 9\times 10^2(-\hat i+3\hat j)$

Hence, $\vec E_1 = (-9\hat i+27\hat j)\times 10^2\,\text{V/m}$

3. Electric field due to $q_2$

Vector from $q_2$ to $P$: $\vec r_2 = (0-4)\hat i + (3-0)\hat j = -4\hat i + 3\hat j$

Magnitude: $r_2 = \sqrt{(-4)^2+3^2}=5$

Now, $\vec E_2 = k\frac{q_2}{r_2^3}\vec r_2$

Substitute: $\vec E_2 = 9\times 10^9\cdot \frac{-25\times 10^{-6}}{5^3}(-4\hat i+3\hat j)$

Since $5^3=125$, $\frac{-25\times 10^{-6}}{125}=-0.2\times 10^{-6}=-2\times 10^{-7}$

Thus, $\vec E_2 = 9\times 10^9\times (-2\times 10^{-7})(-4\hat i+3\hat j)$ $\vec E_2 = -18\times 10^2(-4\hat i+3\hat j)$ $\vec E_2 = (72\hat i-54\hat j)\times 10^2\,\text{V/m}$

4. Net electric field

$\vec E = \vec E_1 + \vec E_2$

$\vec E = (-9\hat i+27\hat j)\times 10^2 + (72\hat i-54\hat j)\times 10^2$

$\vec E = (63\hat i-27\hat j)\times 10^2\,\text{V/m}$

5. Compare with options

The obtained field is $\boxed{\vec E=(63\hat i-27\hat j)\times 10^2\,\text{V/m}}$

This matches Option A.

6. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

So, the stored answer is correct.