Two point charges $q_1\,(\sqrt{10}\ \mu\text{C})$ and $q_2\,(-25\ \mu\text{C})$ are placed on the $x$-axis at $x = 1\ \text{m}$ and $x = 4\ \text{m}$ respectively. The electric field (in V/m) at a point $y = 3\ \text{m}$ on the $y$-axis is $\left[\text{Take } \dfrac{1}{4\pi\epsilon_0} = 9\times10^9\ \text{N m}^2\text{C}^{-2}\right]$
Answer: (D) $\left(63\,\hat i - 27\,\hat j\right)\times10^2$
Use $\vec E = \dfrac{kq\,\vec r}{r^3}$ with $\vec r$ from the charge to $P(0,3)$.
$q_1$: $\vec r = -\hat i + 3\hat j$, $r = \sqrt{10}$, $r^3 = 10\sqrt{10}$
$$\vec E_1 = \frac{9\times10^9\times\sqrt{10}\times10^{-6}}{10\sqrt{10}}(-\hat i+3\hat j) = 900(-\hat i+3\hat j)$$
$q_2$: $\vec r = -4\hat i + 3\hat j$, $r = 5$, $r^3 = 125$
$$\vec E_2 = \frac{9\times10^9\times(-25\times10^{-6})}{125}(-4\hat i+3\hat j) = -1800(-4\hat i+3\hat j) = 7200\,\hat i - 5400\,\hat j$$
$$\vec E = (7200-900)\,\hat i + (2700-5400)\,\hat j = (63\,\hat i - 27\,\hat j)\times10^2\ \text{V/m}$$
Solution by Sreeraj P, M.Sc Physics