The electric flux passing through a surface of area A = 8j m2 in an electric field vector E = 2i + 3j - 4k V/m (bold is for vectors) is:
24 V - m
Electric Flux: Understanding the Concept
Electric flux is a measure of the total number of electric field lines passing through a given surface. It quantifies how much of an electric field passes through a specific area. Electric flux is a scalar quantity, meaning it only has magnitude and no direction. It is a fundamental concept in electromagnetism and is particularly useful in applying Gauss's Law.
For a uniform electric field passing through a flat surface, the electric flux ($\Phi_E$) is calculated using the dot product of the electric field vector ($\mathbf{E}$) and the area vector ($\mathbf{A}$).
Electric Flux Calculation: Step-by-Step
To find the electric flux passing through the given surface, we will use the formula for electric flux, which is the dot product of the electric field vector and the area vector.
Given values:
- Electric field vector, $\mathbf{E} = 2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}$ V/m
- Area vector, $\mathbf{A} = 8\mathbf{j}$ m$^2$
The formula for electric flux ($\Phi_E$) is:
$$\Phi_E = \mathbf{E} \cdot \mathbf{A}$$
Let's substitute the given vectors into the formula:
$$\Phi_E = (2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}) \cdot (0\mathbf{i} + 8\mathbf{j} + 0\mathbf{k})$$
When calculating the dot product of two vectors, we multiply the corresponding components (i with i, j with j, and k with k) and then sum the results.
$$\Phi_E = (2 \times 0) + (3 \times 8) + (-4 \times 0)$$
$$\Phi_E = 0 + 24 + 0$$
$$\Phi_E = 24$$
The unit of electric flux is Volt-meter (V-m), which is derived from the product of the unit of electric field (V/m) and the unit of area (m$^2$).
Therefore, the electric flux passing through the surface is $24$ V-m.
Key Properties of Electric Flux
Understanding electric flux involves several important points:
- Scalar Quantity: Electric flux does not have a direction; it only has magnitude. This is because it is the result of a dot product between two vectors.
- Dependence on Angle: The electric flux through a surface depends on the angle between the electric field lines and the normal to the surface. If the electric field lines are parallel to the surface, the flux is zero. If they are perpendicular, the flux is maximum.
- Gauss's Law Connection: Electric flux is a central component of Gauss's Law, which states that the total electric flux through any closed surface (a Gaussian surface) is proportional to the total electric charge enclosed within that surface.
- Units: The SI unit for electric flux is Newton-meter squared per Coulomb ($\text{N} \cdot \text{m}^2/\text{C}$) or Volt-meter ($\text{V} \cdot \text{m}$). Both units are equivalent.
Summary of Electric Flux Calculation
In this problem, we were given the electric field vector $\mathbf{E} = 2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}$ V/m and the area vector $\mathbf{A} = 8\mathbf{j}$ m$^2$. By applying the definition of electric flux as the dot product of these two vectors, $\Phi_E = \mathbf{E} \cdot \mathbf{A}$, we calculated the value.
The calculation yielded:
| Component | $\mathbf{E}$ Value | $\mathbf{A}$ Value | Product |
|---|---|---|---|
| i | 2 | 0 | $2 \times 0 = 0$ |
| j | 3 | 8 | $3 \times 8 = 24$ |
| k | -4 | 0 | $-4 \times 0 = 0$ |
Summing these products gives the total electric flux: $0 + 24 + 0 = 24$ V-m.
Important Questions from Electric Fields and Gauss' Law
- Two point charges $q_1 \left( {\sqrt {10} {\rm{\mu C}}} \right)$ and $q_2(-18\sqrt{2} {\rm{\mu C}})$ are placed on the x-axis at $x = 0$ m and $x = 4$ m respectively. The electric field (in V/m) at a point $(1, 3)$ m is,
$\left[ {{\rm{Take\;}}\frac{1}{{4{\rm{\pi }}{\epsilon_0}}} = 9 \times {{10}^9}{\rm{N}}{{\rm{m}}^2}{{\rm{C}}^{ - 2}}} \right]$ - Let a total charge $2Q$ be distributed in a sphere of radius $R$, with the charge density given by $\rho(r) = Cr^2$, where $r$ is the distance from the centre. Two charges $A$ and $B$, of $-Q$ each, are placed on diametrically opposite points, at equal distance, '$a$' from the centre. If $A$ and $B$ do not experience any force, then:
The expression for torque '\(\vec{\tau}\)' experienced by an electric dipole of dipole moment '\(\vec{P}\)' in an external uniform electric field '\(\vec{E}\)' is given by :
The surface charge density of a thin spherical shell placed in an air medium is 88.54 c/m2 The intensity of the electric field measured 12 mm outside the shell from the centre of the shell is 5.625 × 101 2 N/C. The thin spherical shell has a radius of:
Which of the following statements is/are correct?
(i) Gauss' law applies to any closed surface, regardless of shape or size.
(ii) We cannot distinguish between positive and negative flux depending on the direction of the electric flux lines.
(iii) The net electric flux leaving a surface will always be zero if there is a charge bound inside of it.
