Electric Field - Electric Field Intensity Due to a Point-Charge

Step 1: Consider a source point charge q placed at the origin, and a test charge q₀ placed at point P, at a distance r.

Step 2: By Coulomb's Law, the force on q₀​ due to q is:

Step 3: Electric field intensity at P is force per unit test charge:

Step 4: For a dielectric medium with dielectric constant K:

When several point charges act on a point, the net field is the vector sum of individual fields:

\[\vec{E}=\vec{E}_1+\vec{E}_2+\cdots+\vec{E}_n=\sum_{i=1}^n\frac{1}{4\pi\varepsilon_0}\frac{q_i}{r_i^2}\hat{r}_i\]

Real-Life Analogy: Think of each charge as a speaker emitting sound in all directions. At any listening point, you hear the combined effect of all speakers — similarly, a point in space experiences the vector sum of all individual electric fields.

Problem: A charge of −2 μC is located at point A(2, 2, 2) m. Find the electric field intensity at point B(1, 1, 1) m.

Solution:

1. Displacement vector: \[\vec r\] = B − A = (−1,−1,−1) m, so r = \[\sqrt 3\] m

2. Unit vector: \[\hat r\] = \[\frac{-(\hat{i}+\hat{j}+\hat{k})}{\sqrt3}\]

3. Applying the formula:
   \[\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{r}=9\times10^9\times\frac{-2\times10^{-6}}{3}\times\frac{-(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}\]

4. Result: \[\vec{E}=2000\sqrt{3}\left[-(\hat{i}+\hat{j}+\hat{k})\right]\mathrm{NC}^{-1}\]

• Electric field intensity: \[\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{r}\].
• It is a vector quantity, directed outward for +q and inward for −q.
• Follows an inverse-square law with distance.
• Independent of the test charge used to measure it.
• For multiple charges, use vector superposition.
• In a medium, divide by dielectric constant K.