krit.club logo

Electric Charges and Fields - Electric Flux

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Electric flux ΦE\Phi_E is defined as the total number of electric field lines passing through a given area held inside an electric field.

•

The area vector A⃗\vec{A} of a surface is a vector whose magnitude is equal to the surface area and whose direction is perpendicular (normal) to the surface. For a closed surface, the area vector is always directed outwards.

•

Electric flux is a scalar quantity. Its SI unit is N⋅m2⋅C−1N \cdot m^2 \cdot C^{-1} or Volt-meter (V⋅mV \cdot m).

•

The flux through a surface is maximum when the electric field E⃗\vec{E} is perpendicular to the surface (theta=0∘\\theta = 0^\circ relative to the normal).

•

The flux through a surface is zero when the electric field E⃗\vec{E} is parallel to the surface (θ=90∘\theta = 90^\circ relative to the normal).

•

If the angle between E⃗\vec{E} and the normal to the area is acute (0≤θ<90∘0 \le \theta < 90^\circ), the flux is positive (outward flux). If it is obtuse (90∘<θ≤180∘90^\circ < \theta \le 180^\circ), the flux is negative (inward flux).

📐Formulae

ΦE=E⃗⋅A⃗=EAcos⁡θ\Phi_E = \vec{E} \cdot \vec{A} = EA \cos \theta

ΦE=∫SE⃗⋅dA⃗\Phi_E = \int_S \vec{E} \cdot d\vec{A}

Φtotal=∮SE⃗⋅dA⃗=qenclosedϵ0\Phi_{total} = \oint_S \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}

[ΦE]=[M1L3T−3A−1][\Phi_E] = [M^1 L^3 T^{-3} A^{-1}]

💡Examples

Problem 1:

A square surface of side L=10 cmL = 10\text{ cm} is placed in a uniform electric field E⃗=3×103i^ N/C\vec{E} = 3 \times 10^3 \hat{i}\text{ N/C}. Calculate the flux through the square if the normal to its plane makes a 60∘60^\circ angle with the X-axis.

Solution:

Given: E=3×103 N/CE = 3 \times 10^3\text{ N/C}, Side L=0.1 mL = 0.1\text{ m}, θ=60∘\theta = 60^\circ. Area A=L2=(0.1)2=0.01 m2A = L^2 = (0.1)^2 = 0.01\text{ m}^2. Using ΦE=EAcos⁡θ\Phi_E = EA \cos \theta: ΦE=(3×103)×(0.01)×cos⁡(60∘)\Phi_E = (3 \times 10^3) \times (0.01) \times \cos(60^\circ) ΦE=30×0.5=15 N⋅m2/C\Phi_E = 30 \times 0.5 = 15\text{ N}\cdot\text{m}^2/C.

Explanation:

The flux is calculated by taking the dot product of the electric field and the area vector. Since the normal to the plane (the area vector direction) is already given as 60∘60^\circ to the field, we use θ=60∘\theta = 60^\circ directly in the cosine function.

Problem 2:

If a point charge q=+2 \muCq = +2\text{ \mu C} is placed at the center of a cubic Gaussian surface of side 9 cm9\text{ cm}, what is the net electric flux through the surface?

Solution:

By Gauss's Law, Φtotal=qϵ0\Phi_{total} = \frac{q}{\epsilon_0}. Given q=2×10−6 Cq = 2 \times 10^{-6}\text{ C} and ϵ0≈8.854×10−12 C2 N−1 m−2\epsilon_0 \approx 8.854 \times 10^{-12}\text{ C}^2\text{ N}^{-1}\text{ m}^{-2}. Φtotal=2×10−68.854×10−12≈2.26×105 N⋅m2/C\Phi_{total} = \frac{2 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 2.26 \times 10^5\text{ N}\cdot\text{m}^2/C.

Explanation:

According to Gauss's Law, the total electric flux through a closed surface depends only on the net charge enclosed and is independent of the shape or size of the surface.

Electric Flux - Revision Notes & Key Formulas | CBSE Class 12 Physics