A cone of base radius R and height h is located in a uniform electric field $\overrightarrow{E}$ parallel to its base. The electric flux entering the cone is :
- $\frac{1}{2}$ E h R
- E h R
- 2 E h R
- 4 E h R
The Correct Option is B
Solution and Explanation
$= E\left(\frac{1}{2}h\times2R\right)$
$= Ehr$
Top Questions on Gauss Law
- If the net flux through a cube is 1.05 N m\(^2\) C\(^{-1}\), what will be the total charge inside the cube? (Given: The permittivity of free space is \(8.85 \times 10^{-12}\) C\(^2\) N\(^{-1}\) m\(^{-2}\)).
- CUET (UG) - 2025
- Physics
- Gauss Law
- A charge q is placed at the center of one of the surface of a cube. The flux linked with the cube is :-
- JEE Main - 2024
- Physics
- Gauss Law
- An electric field \( \vec{E} = (2x \hat{i}) \, \text{N C}^{-1} \) exists in space. A cube of side \( 2 \, \text{m} \) is placed in the space as per the figure given below. The electric flux through the cube is __________ \( \text{N m}^2/\text{C} \).
- JEE Main - 2024
- Physics
- Gauss Law
- There are two cubical Gaussian surface carrying charges as shown. Find ratio of fluxes through surface \(C_1\) and \(C_2\):
- JEE Main - 2024
- Physics
- Gauss Law
- Match List I with List IIChoose the correct answer from the options given below:
LIST I LIST II A Gauss's Law in Electrostatics I \(\oint \vec{E} \cdot d \vec{l}=-\frac{d \phi_B}{d t}\) B Faraday's Law II \(\oint \vec{B} \cdot d \vec{A}=0\) C Gauss's Law in Magnetism III \(\oint \vec{B} \cdot d \vec{l}=\mu_0 i_c+\mu_0 \in_0 \frac{d \phi_E}{d t}\) D Ampere-Maxwell Law IV \(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\) - JEE Main - 2023
- Physics
- Gauss Law
Questions Asked in JEE Main exam
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]- JEE Main - 2026
- Ray optics and optical instruments
A small block of mass \(m\) slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration \(a_0\). The angle between the inclined plane and ground is \(\theta\) and its base length is \(L\). Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is _______.
- JEE Main - 2026
- General Physics
- Let \(\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}\), \(\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}\) and \(\vec{c}\) be a vector such that \(\vec{a} \times \vec{c} = \vec{b}\). If \(\vec{c} \cdot (\hat{i}+\hat{j}+\hat{k}) = 4\), then \(|\vec{a}+\vec{c}|^2\) is equal to:
- JEE Main - 2026
- Vector Algebra
- Let \( \alpha \) and \( \beta \) respectively be the maximum and the minimum values of the function \( f(\theta) = 4\left(\sin^4\left(\frac{7\pi}{2} - \theta\right) + \sin^4(11\pi + \theta)\right) - 2\left(\sin^6\left(\frac{3\pi}{2} - \theta\right) + \sin^6(9\pi - \theta)\right) \). Then \( \alpha + 2\beta \) is equal to :
- JEE Main - 2026
- Statistics
- The crystal field splitting energy of $[Co(oxalate)_3]^3-$ complex is 'n' times that of the $[Cr(oxalate)_3]^3-$ complex. Here 'n' is_______ (Assume $\Delta_0 > P$)}
- JEE Main - 2026
- Solutions
Concepts Used:
Gauss Law
Gauss law states that the total amount of electric flux passing through any closed surface is directly proportional to the enclosed electric charge.
Gauss Law:
According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.
For example, a point charge q is placed inside a cube of edge ‘a’. Now as per Gauss law, the flux through each face of the cube is q/6ε0.
Gauss Law Formula:
As per the Gauss theorem, the total charge enclosed in a closed surface is proportional to the total flux enclosed by the surface. Therefore, if ϕ is total flux and ϵ0 is electric constant, the total electric charge Q enclosed by the surface is;
Q = ϕ ϵ0
The Gauss law formula is expressed by;
ϕ = Q/ϵ0
Where,
Q = total charge within the given surface,
ε0 = the electric constant.