গাউস থেইকা কুলম্ব

Q1(a):
State Gauss’s Law in electrostatics and explain its physical significance. Then, using a suitable Gaussian surface and symmetry, derive Coulomb’s Law for the electric field intensity due to a point charge from Gauss’s Law. Include all necessary mathematical steps and assumptions.

✅ Structured Answer:¶

🔷 1. Gauss’s Law – Statement¶

Gauss’s Law states that:
“The total electric flux through any closed surface is equal to the net electric charge enclosed by that surface divided by the permittivity of free space.”

🔸 Mathematical Form:¶

\[ \boxed{\oint \vec{E} \cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}} \]

Where:

\(\vec{E}\) = Electric field vector
\(d\vec{A}\) = Infinitesimal area vector on closed surface
\(q_{\text{enc}}\) = Net charge enclosed inside the surface
\(\varepsilon_0\) = Permittivity of free space \((8.854 \times 10^{-12} \, \text{F/m})\)

🔷 2. Physical Significance of Gauss’s Law¶

Relates electric field and charge distribution
Useful in calculating electric field when there is high symmetry (spherical, cylindrical, planar)
A fundamental law in Maxwell’s equations

🔷 3. Deduction of Coulomb’s Law from Gauss’s Law¶

We will now derive Coulomb’s Law (electric field due to a point charge) using Gauss’s Law.

✅ Assumptions:¶

Let a point charge \(q\) be placed at the origin.
Choose a spherical Gaussian surface of radius \(r\) centered at the point charge.

✅ Step 1: Apply Gauss’s Law¶

\[ \oint \vec{E} \cdot d\vec{A} = \frac{q}{\varepsilon_0} \]

Due to spherical symmetry:

\(\vec{E}\) is constant in magnitude over the surface
\(\vec{E} \parallel d\vec{A}\), so \(\vec{E} \cdot d\vec{A} = E\,dA\)

\[ E \oint dA = \frac{q}{\varepsilon_0} \]

✅ Step 2: Evaluate Surface Area¶

Total surface area of a sphere of radius \(r\):

\[ \oint dA = 4\pi r^2 \]

Substitute:

\[ E \cdot 4\pi r^2 = \frac{q}{\varepsilon_0} \]

✅ Step 3: Solve for \(E\)¶

\[ \boxed{E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2}} \]

This is Coulomb’s Law for electric field intensity due to a point charge.

🔷 4. Vector Form of Electric Field¶

\[ \vec{E} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{r^2} \cdot \hat{r} \]

Where \(\hat{r}\) is the unit vector pointing radially outward from the charge (if \(q > 0\)).

🔷 5. From Field to Force (Coulomb’s Force)¶

Force on a test charge \(q_0\):

\[ \vec{F} = q_0 \vec{E} = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q q_0}{r^2} \cdot \hat{r} \]

This is the standard form of Coulomb’s Law.

✅ Conclusion:¶

Gauss’s Law provides a powerful and elegant method for evaluating electric fields using symmetry and surface integrals.
By applying it to a spherical surface around a point charge, we easily deduce Coulomb’s Law, which was historically discovered first but can now be derived from more fundamental laws.

Let me know if you’d like a diagram to visualize the derivation, or a Bangla explanation for better understanding.