Numerical Problems Based on Electric Flux for Class 12 Physics

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Home CBSE Class 12 Physics Numerical Problems Numerical Problems Based on Electric Flux for Class 12 Physics

Electric flux

The term flux implies some kind of flow. Flux is the property of any vector field. The electric flux is a property of electric field.

The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area.

∆ϕE = E ∆S

Note:

  • Electric flux is a scalar quantity.
  • Unit of ϕE = Unit of E × unit of S
  • ∴ SI unit of electric flux = NC-1.m2 = Nm2C_1.
  • Equivalently, SI unit of electric flux = Vm_1.m2 = Vm.

Gauss’s theorem.

This theorem gives a relationship between the total flux passing through any closed surface and the net charge enclosed within the surface.

Gauss theorem states that the total flux through a closed surface is 1/ε0 times the net charge enclosed by the closed surface.

Mathematically, it can be expressed as

$$
\phi_E=\oint_5 \vec{E} \cdot \overrightarrow{d S}=\frac{q}{\varepsilon_0}
$$

Points to Note:

  • Gauss’s theorem is valid for a closed surface of any shape and for any general charge distribution.
  • If the net charge enclosed by a closed surface is zero (q = 0), then flux through it is also zero.
  • The net flux through a closed surface due to a charge lying outside the closed surface is zero.
  • The charge q appearing in the Gauss’s theorem includes the sum of all the charges located anywhere inside the closed surface.

Numerical Problems on Electric Flux

Class 12 Physics · Electric Charges and Fields
SQ

Electric Flux Practice Set

Apply $\Phi = \vec{E} \cdot \vec{A}$ and Gauss’s Law
Ch 1 · Electrostatics
1
If the electric field is given by $\vec{E} = 8\hat{i} + 4\hat{j} + 3\hat{k}\text{ NC}^{-1}$, calculate the electric flux through a surface of area $100\text{ m}^2$ lying in the X-Y plane.
Answer $300\text{ Nm}^2\text{C}^{-1}$ 📝
Detailed Solution

Since the area lies in the X-Y plane, its area vector $\vec{A}$ acts along the normal to the X-Y plane, which is the Z-axis (denoted by the unit vector $\hat{k}$).

Area vector, $\vec{A} = 100\hat{k}\text{ m}^2$.

Electric flux ($\Phi$) is the dot product of the electric field and the area vector:

$$\Phi = \vec{E} \cdot \vec{A} = (8\hat{i} + 4\hat{j} + 3\hat{k}) \cdot (100\hat{k})$$
$$\Phi = (8 \times 0) + (4 \times 0) + (3 \times 100) = 300\text{ Nm}^2\text{C}^{-1}$$
2
The electric field in a certain region of space is $(5\hat{i} + 4\hat{j} – 4\hat{k}) \times 10^5\text{ NC}^{-1}$. Calculate the electric flux due to this field over an area of $(2\hat{i} – \hat{j}) \times 10^{-2}\text{ m}^2$.
Answer $6 \times 10^3\text{ Nm}^2\text{C}^{-1}$ 📝
Detailed Solution

Given Electric Field $\vec{E} = (5\hat{i} + 4\hat{j} – 4\hat{k}) \times 10^5\text{ NC}^{-1}$.

Given Area vector $\vec{A} = (2\hat{i} – \hat{j} + 0\hat{k}) \times 10^{-2}\text{ m}^2$.

Electric Flux ($\Phi$):

$$\Phi = \vec{E} \cdot \vec{A} = \left[ (5\hat{i} + 4\hat{j} – 4\hat{k}) \times 10^5 \right] \cdot \left[ (2\hat{i} – \hat{j}) \times 10^{-2} \right]$$
$$\Phi = [(5 \times 2) + (4 \times -1) + (-4 \times 0)] \times 10^5 \times 10^{-2}$$
$$\Phi = [10 – 4] \times 10^3 = 6 \times 10^3\text{ Nm}^2\text{C}^{-1}$$
3
Consider a uniform electric field $\vec{E} = 3 \times 10^3\hat{i}\text{ NC}^{-1}$. Calculate the flux of this field through a square surface of area $100\text{ cm}^2$ (a square of $10\text{ cm}$ side) when (i) its plane is parallel to the y-z plane, and (ii) the normal to its plane makes a $60^\circ$ angle with the x-axis.
Answer (i) $30\text{ Nm}^2\text{C}^{-1}$ (ii) $15\text{ Nm}^2\text{C}^{-1}$ 📝
Detailed Solution

First, convert the area to SI units: $A = 100\text{ cm}^2 = 100 \times 10^{-4}\text{ m}^2 = 10^{-2}\text{ m}^2$. The magnitude of the electric field is $E = 3 \times 10^3\text{ NC}^{-1}$ pointing along the X-axis.

(i) Plane parallel to the y-z plane:

When the surface is parallel to the Y-Z plane, its normal vector points directly along the X-axis. Therefore, the angle $\theta$ between the normal and the electric field (which is also along the X-axis) is $0^\circ$.

$$\Phi_1 = EA\cos\theta = (3 \times 10^3) \times (10^{-2}) \times \cos 0^\circ$$
$$\Phi_1 = 30 \times 1 = 30\text{ Nm}^2\text{C}^{-1}$$

(ii) Normal makes a $60^\circ$ angle with the x-axis:

Here, the angle $\theta$ between the area vector (normal) and the electric field is $60^\circ$.

$$\Phi_2 = EA\cos\theta = (3 \times 10^3) \times (10^{-2}) \times \cos 60^\circ$$
$$\Phi_2 = 30 \times 0.5 = 15\text{ Nm}^2\text{C}^{-1}$$
4
Given a uniform electric field $\vec{E} = 5 \times 10^3\hat{i}\text{ NC}^{-1}$, find the flux of this field through a square of $10\text{ cm}$ on a side whose plane is parallel to the Y-Z plane. What would be the flux through the same square if the plane makes a $30^\circ$ angle with the X-axis?
Answer (i) $50\text{ Nm}^2\text{C}^{-1}$ (ii) $25\text{ Nm}^2\text{C}^{-1}$ 📝
Detailed Solution

Area of the square $A = (10\text{ cm})^2 = 100\text{ cm}^2 = 10^{-2}\text{ m}^2$. The field is $E = 5 \times 10^3\text{ NC}^{-1}$ along the X-axis.

(i) Plane parallel to the Y-Z plane:

The normal vector is along the X-axis, meaning $\theta = 0^\circ$.

$$\Phi_1 = EA\cos 0^\circ = (5 \times 10^3) \times (10^{-2}) \times 1 = 50\text{ Nm}^2\text{C}^{-1}$$

(ii) Plane makes a $30^\circ$ angle with the X-axis:

The angle is given between the plane and the X-axis. The angle $\theta$ required for the flux formula is the angle between the normal to the plane and the X-axis. Therefore, $\theta = 90^\circ – 30^\circ = 60^\circ$.

$$\Phi_2 = EA\cos 60^\circ = (5 \times 10^3) \times (10^{-2}) \times \frac{1}{2}$$
$$\Phi_2 = 50 \times 0.5 = 25\text{ Nm}^2\text{C}^{-1}$$
7

A charge $q$ is situated at the centre of an imaginary hemispherical surface, as shown in Fig. 1.93. Using Gauss’s theorem and symmetry considerations, determine the electric flux due to this charge through hemispherical surface.

Answer $\frac{q}{2\varepsilon_0}$ 📝
Detailed Solution

According to Gauss’s Theorem, the total electric flux through any closed Gaussian surface enclosing a charge $q$ is equal to $\frac{q}{\varepsilon_0}$.

To use symmetry, imagine completing the hemisphere into a full, closed spherical surface with the charge $q$ exactly at its center. The total flux through this entire sphere would be:

$$\Phi_{total} = \frac{q}{\varepsilon_0}$$

Because the charge is perfectly centered, the electric field lines are distributed symmetrically in all directions. Therefore, exactly half of the total flux will pass through the upper hemisphere.

$$\Phi_{hemisphere} = \frac{1}{2}\Phi_{total} = \frac{q}{2\varepsilon_0}$$

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