Two point charges $$q_1(\sqrt{10}$$ $$\mu$$C) and $$q_2(-25$$ $$\mu$$C) are placed on the $$x$$-axis at $$x = 1$$ m and $$x = 4$$ m respectively. The electric field (in V/m) at a point $$y = 3$$ m on $$y$$-axis is, [Take $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$ N m$$^2$$C$$^{-2}$$]

We have two point charges on the $$x$$-axis. The first charge is $$q_1 = +\sqrt{10}\;\mu{\rm C}=+\sqrt{10}\times10^{-6}\;{\rm C}$$ located at $$x=1\;{\rm m},\,y=0\;{\rm m}$$. The second charge is $$q_2 = -25\;\mu{\rm C}=-25\times10^{-6}\;{\rm C}$$ placed at $$x=4\;{\rm m},\,y=0\;{\rm m}$$. The point where we want the electric field is on the $$y$$-axis at $$x=0,\,y=3\;{\rm m}$$.

The electric field due to a point charge is given by the formula

$$\vec E=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}\,\hat r,$$

where $$\hat r$$ is the unit vector drawn from the charge towards the field point and $$r$$ is the distance between them. We shall compute this field separately for each charge and then add the two vector results.

Field due to $$q_1$$

The position vector from $$q_1(1,0)$$ to the point $$(0,3)$$ is

$$\vec r_1 = (0-1)\hat i + (3-0)\hat j = -1\hat i + 3\hat j.$$

The magnitude of this vector is

$$r_1=\sqrt{(-1)^2+3^2}=\sqrt{1+9}=\sqrt{10}\;{\rm m}.$$

The unit vector is therefore

$$\hat r_1=\dfrac{\vec r_1}{r_1}= \dfrac{-1}{\sqrt{10}}\hat i+\dfrac{3}{\sqrt{10}}\hat j.$$

Now the magnitude of the field is

$$E_1=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_1|}{r_1^2} = 9\times10^9 \times \dfrac{\sqrt{10}\times10^{-6}}{(\sqrt{10})^2} = 9\times10^9 \times \dfrac{\sqrt{10}\times10^{-6}}{10} =9\times10^2\sqrt{10}\;{\rm V/m}.$$

Multiplying this magnitude with the unit vector, we obtain

$$\vec E_1 = 9\times10^2\sqrt{10}\left(\dfrac{-1}{\sqrt{10}}\hat i+\dfrac{3}{\sqrt{10}}\hat j\right) = 900(-1\hat i+3\hat j) = -900\hat i + 2700\hat j\;{\rm V/m}.$$

Field due to $$q_2$$

The position vector from $$q_2(4,0)$$ to the point $$(0,3)$$ is

$$\vec r_2 = (0-4)\hat i + (3-0)\hat j = -4\hat i + 3\hat j.$$

The magnitude of this vector is

$$r_2=\sqrt{(-4)^2+3^2}=\sqrt{16+9}=5\;{\rm m}.$$

Thus the unit vector is

$$\hat r_2=\dfrac{\vec r_2}{r_2}= \dfrac{-4}{5}\hat i+\dfrac{3}{5}\hat j.$$

The magnitude of the field that a charge of magnitude $$|q_2|$$ would produce is

$$E_2=\dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_2|}{r_2^2} =9\times10^9 \times \dfrac{25\times10^{-6}}{5^2} =9\times10^9 \times \dfrac{25\times10^{-6}}{25} =9\times10^3\;{\rm V/m}=9000\;{\rm V/m}.$$

Because $$q_2$$ is –ve, the field vector points towards the charge, i.e. opposite to $$\hat r_2$$. Hence

$$\vec E_2 = -E_2\,\hat r_2 = -9000\left(\dfrac{-4}{5}\hat i+\dfrac{3}{5}\hat j\right) = 9000\left(\dfrac{4}{5}\hat i-\dfrac{3}{5}\hat j\right) = 9000\!\left(0.8\hat i-0.6\hat j\right) = 7200\hat i - 5400\hat j\;{\rm V/m}.$$

Net electric field

Now we add the two vectors:

$$\vec E = \vec E_1 + \vec E_2 = (-900\hat i + 2700\hat j) + (7200\hat i - 5400\hat j) = ( -900 + 7200)\hat i + (2700 - 5400)\hat j = 6300\hat i - 2700\hat j\;{\rm V/m}.$$

Factoring out $$10^2$$, we write

$$\vec E = (63\hat i - 27\hat j)\times10^2\;{\rm V/m}.$$

Among the given alternatives this matches option D.

Hence, the correct answer is Option D.