The electric field due to an infinitely long straight uniformly charged wire at a distance $r$ is directly proportional to
- r
- $r^{2}$
- $\frac{1}{r}$
- $\frac{1}{r^{2}}$
The Correct Option is C
Solution and Explanation
$E=\frac{\lambda}{2\pi\varepsilon_{0} r}$
where $\lambda$ is the uniform linear charge density and $\varepsilon_{0}$ is the permittivity of free space.
$\therefore\, E \propto \frac{1}{r}$
Top Questions on 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
A cubical volume is bounded by the surfaces \( x = 0 \), \( x = a \), \( y = 0 \), \( y = a \), \( z = 0 \), \( z = a \). The electric field in the region is given by: \[ \vec{E} = E_0 x \hat{i} \]
Where \( E_0 = 4 \times 10^4 \, \text{NC}^{-1} \, \text{m}^{-1} \). If \( a = 2 \, \text{cm} \), the charge contained in the cubical volume is \( Q \times 10^{-14} \, \text{C} \). The value of \( Q \) is ______. (Take \( \epsilon_0 = 9 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \))- JEE Main - 2023
- Physics
- Gauss Law
Questions Asked in KEAM exam
- Let
\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]
Then \( f'(-4) \) is equal to:- KEAM - 2024
- introduction to three dimensional geometry
If \( f'(x) = 4x\cos^2(x) \sin\left(\frac{x}{4}\right) \), then \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:
- KEAM - 2024
- introduction to three dimensional geometry
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
- KEAM - 2024
- Magnitude and Directions of a Vector
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
- KEAM - 2024
- Magnitude and Directions of a Vector
- Evaluate the integral:
\[ \int \frac{4x \cos \left( \sqrt{4x^2 + 7} \right)}{\sqrt{4x^2 + 7}} \, dx \]
- KEAM - 2024
- Integral Calculus
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.