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 __________ \( \text{N m}^2/\text{C} \).
Correct Answer: 16
Approach Solution - 1
The electric flux (Φ) through a closed surface such as a cube can be calculated using Gauss's law: \( Φ = \oint \vec{E} \cdot d\vec{A} \), where \( \vec{E} \) is the electric field, and \( d\vec{A} \) is the differential area vector. Given the electric field \( \vec{E} = (2x \hat{i}) \, \text{N C}^{-1} \), the flux through surfaces perpendicular to other axes will be zero because the field is only in the x-direction.
Consider the cube positioned from \( x = 0 \) to \( x = 2 \) with side length \( a = 2 \, \text{m} \). The cube has two faces perpendicular to the x-axis at \( x = 0 \) and \( x = 2 \). The area \( A \) of each face is \( A = 2 \times 2 = 4 \, \text{m}^2 \).
For the face at \( x = 0 \):
\( \vec{E} = 2(0) \hat{i} = 0 \) so flux, \( Φ_0 = E \cdot A = 0 \times 4 = 0 \, \text{N m}^2/\text{C} \).
For the face at \( x = 2 \):
\( \vec{E} = 2(2) \hat{i} = 4 \, \text{N C}^{-1} \) and outward \( d\vec{A} = 4 \hat{i} \), so flux, \( Φ_2 = E \cdot A = 4 \times 4 = 16 \, \text{N m}^2/\text{C} \).
The total flux through the cube is \( Φ_{\text{total}} = Φ_0 + Φ_2 = 0 + 16 = 16 \, \text{N m}^2/\text{C} \).
The computed value \( 16 \, \text{N m}^2/\text{C} \) falls within the expected range of (16,16).
Approach Solution -2
The electric flux is given by Gauss's Law:
\(\Phi = \oint \vec{E} \cdot d\vec{A}.\)
The field $\vec{E} = 2x \hat{i}$ varies with $x$.
For the cube, only the left ($x = 0$) and right ($x = 2$) faces contribute: \(\Phi = E_\text{right} A - E_\text{left} A.\)
Substituting $A = 4 \, \mathrm{m}^2$, $E_\text{right} = 2(2) = 4$, and $E_\text{left} = 2(0) = 0$:
\(\Phi = 4 \cdot 4 - 0 \cdot 4 = 16 \, \mathrm{Nm}^2/\mathrm{C}.\)
Top JEE Main Physics Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,
- $I _{ CM }$ is the moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc $I_{A B}$ is it's moment of inertia about an axis $AB$ perpendicular to plane and parallel to axis $CM$ at a distance $\frac{2}{3} R$ from center Where $R$ is the radius of the dise The ratio of $I _{ AB }$ and $I _{ CM }$ is $x: 9$ The value of $x$ is______
- A nucleus disintegrates into two smaller parts, which have their velocities in the ratio $3: 2$ The ratio of their nuclear sizes will be $\left(\frac{x}{3}\right)^{\frac{1}{3}}$ The value of ' $x$ ' is:-
Top JEE Main Gauss Law Questions
- There are two cubical Gaussian surface carrying charges as shown. Find ratio of fluxes through surface \(C_1\) and \(C_2\):
- A charge q is placed at the center of one of the surface of a cube. The flux linked with the cube is :-
- 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 \(\oint \vec{E} \cdot d \vec{s}=\frac{q}{\epsilon_0}\) 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 \( \epsilon_0 = 9 \times 10^{-12} \, \text{C}^2/\text{Nm}^2 \))- The electric potential at the center of two concentric half rings of radii $R_1$ and $R_2$, having same linear charge density $\lambda$ is :
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.