Animated Solution for Physics - Electrostatics: 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 Y-axis is $\left(\text{Take, }\frac{1}{4\pi {ϵ}_0}=9\times 10^9{\text{ N-m}}^2{\text{C}}^{-2}\right)$

Imagine a coordinate plane as a battlefield where electric charges exert their influence. We have a positive charge $q_1=\sqrt{10}\mu \text{C}$ stationed at point $A(1,0)$, and a negative charge $q_2=-25\mu \text{C}$ holding the fort at point $B(4,0)$. Our mission is to determine the net electric field at a specific target: point $C(0,3)$ on the y-axis.

To conquer this problem, we must rely on the Principle of Superposition. This principle states that the total electric field at any point is simply the vector sum of the individual electric fields created by each charge independently.

Let's focus entirely on $q_1$ first. To find the electric field it creates at point $C$, we need the distance between them. Looking at the coordinates, the path from $A(1,0)$ to $C(0,3)$ forms a right-angled triangle with the axes.

Using the Pythagorean theorem, the distance $AC$ is:

Now, we calculate the magnitude of the electric field ${\mathbf{E}}_1$ using Coulomb's Law:

Substituting the given values:

Because $q_1$ is positive, the electric field ${\mathbf{E}}_1$ points away from it. Geometrically, this means it points to the left (negative x-direction) and upwards (positive y-direction). Let ${\theta }_1$ be the angle inside the triangle at vertex $A$. We can resolve ${\mathbf{E}}_1$ into its components:

Now, we shift our attention to $q_2$. The distance from $B(4,0)$ to $C(0,3)$ forms another right-angled triangle. This one is a classic 3-4-5 triangle!

Let's calculate the magnitude of ${\mathbf{E}}_2$. Remember, when calculating magnitude, we only use the absolute value of the charge:

Notice how beautifully the $25$ cancels out!

Since $q_2$ is a negative charge, the electric field ${\mathbf{E}}_2$ points towards it. This means it points to the right (positive x-direction) and downwards (negative y-direction). Let ${\theta }_2$ be the angle at vertex $B$:

Resolving ${\mathbf{E}}_2$ into components:

We have successfully broken down the complex fields into manageable $\hat{i}$ and $\hat{j}$ components. The final step is to simply add them together to find the net electric field ${\mathbf{E}}_{net}$:

Grouping the like terms:

And there we have it! By systematically breaking down the geometry, calculating magnitudes, and carefully assigning vector directions based on the signs of the charges, we arrive at the perfect solution.