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.
Only (i)
Gauss' Law Explained
Gauss' law is a fundamental principle in electromagnetism that relates the electric flux through a closed surface to the net electric charge enclosed within that surface. It is a powerful tool for calculating electric fields, especially in situations with high degrees of symmetry.
Electric Flux and its Direction
Electric flux is a measure of the electric field passing through a given surface. The direction of electric flux lines helps us understand whether the flux is positive or negative. Let's analyze each statement regarding Gauss' law and electric flux:
- Statement (i): Gauss' law applies to any closed surface, regardless of shape or size.
This statement is correct. Gauss' law is a universal law that holds true for any arbitrary closed surface, often called a Gaussian surface. The validity of the law does not depend on the shape (spherical, cylindrical, irregular, etc.) or the size of this closed surface. The key requirement is that the surface must be closed, meaning it encloses a volume, and the law relates the total flux through this entire closed surface to the net charge enclosed within that volume.
Mathematically, Gauss' law is expressed as:
\( \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} \)
Where:
- \( \Phi_E \) is the total electric flux through the closed surface.
- \( \vec{E} \) is the electric field.
- \( d\vec{A} \) is a differential area vector element on the surface, pointing outwards.
- \( Q_{enc} \) is the net electric charge enclosed by the surface.
- \( \epsilon_0 \) is the permittivity of free space.
- Statement (ii): We cannot distinguish between positive and negative flux depending on the direction of the electric flux lines.
This statement is incorrect. We absolutely can distinguish between positive and negative electric flux based on the direction of the electric flux lines relative to the closed surface. By convention:
- Positive flux occurs when electric field lines emerge from (or leave) the closed surface. This typically happens when there is a net positive charge enclosed within the surface.
- Negative flux occurs when electric field lines enter (or go into) the closed surface. This typically happens when there is a net negative charge enclosed within the surface.
The dot product \( \vec{E} \cdot d\vec{A} \) in the integral definition of flux inherently accounts for this directionality. If \( \vec{E} \) and \( d\vec{A} \) (which points outwards) are in the same general direction, the flux is positive. If they are in opposite directions, the flux is negative.
- Statement (iii): The net electric flux leaving a surface will always be zero if there is a charge bound inside of it.
This statement is incorrect. According to Gauss' law, the net electric flux leaving a closed surface is directly proportional to the net charge enclosed within it, i.e., \( \Phi_E = \frac{Q_{enc}}{\epsilon_0} \). Therefore, if there is a net charge \( Q_{enc} \) bound inside the surface (meaning \( Q_{enc} \neq 0 \)), the net electric flux leaving the surface will not be zero. It will only be zero if the net charge enclosed is zero (e.g., if an equal amount of positive and negative charges are enclosed, making \( Q_{enc} = 0 \), or if there is no charge inside).
Conclusion on Statements
Based on the analysis:
- Statement (i) is correct.
- Statement (ii) is incorrect.
- Statement (iii) is incorrect.
Therefore, only statement (i) is correct.
Important Questions from Electric Fields and Gauss' Law
A positive charge +q is placed at the centre of a hollow metallic sphere of inner radius a and outer radius b. the electric field at a distance r from the centre is denoted by E. In this regards, which one of the following statement is correct?
If a free electron moves through a potential difference of 1 kV, then the energy gained by the electron is given by
- 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 :
