Question:
An electric field is given by \( (6\hat{i} + 5\hat{j} + 3\hat{k}) \, \text{N/C} \). The electric flux through a surface area \( 30\hat{i} \, \text{m}^2 \) lying in the YZ-plane (in SI units) is:
An electric field is given by \( (6\hat{i} + 5\hat{j} + 3\hat{k}) \, \text{N/C} \). The electric flux through a surface area \( 30\hat{i} \, \text{m}^2 \) lying in the YZ-plane (in SI units) is:
Updated On: Nov 10, 2024
- 90
- 150
- 180
- 60
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
The Correct Answer is
Was this answer helpful?
0
0
Top Questions on Electric Flux Density
- An electric field, \(\vec{E} = \frac{2\hat{i} + 6\hat{j} + 8\hat{k}}{\sqrt{6}}\) passes through the surface of \(4 \, \text{m}^2\) area having unit vector \(\hat{n} = \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}\). The electric flux for that surface is _________ \( \text{V m} \).
- JEE Main - 2024
- Physics
- Electric Flux Density
- Given below are two statements:
Statement-I: The tangential component of the electric field intensity is continuous across an interface between two general materials, regardless of charge densities on the surface.
Statement-II The tangential component of the electric flux density is discontinuous across the interface. The discontinuity is equal to the ratio between the permittivities of the materials.
In the light of the above statements, choose the correct answer from the options given below.- CUET (PG) - 2023
- Electromagnetic Fields
- Electric Flux Density
- In a Bainbridge mass spectrograph singly ionized atoms of a Neon-20 pass into the deflection chamber with the velocity of 105 m/sec. If they are deflected by a magnetic field of flux density 0.08 tesla, then the path radius of Neon-20 ion is ____
- SRMJEEE - 2018
- Physics
- Electric Flux Density
Questions Asked in JEE Main exam
- A convex lens of focal length 40 cm forms an image of an extended source of light on a photo-electric cell. A current \( I \) is produced. The lens is replaced by another convex lens having the same diameter but focal length 20 cm. The photoelectric current now is:
- JEE Main - 2024
- Dual nature of matter
- For a sparingly soluble salt \( \text{AB}_2 \), the equilibrium concentrations of \( \text{A}^{2+} \) ions and \( \text{B}^- \) ions are \( 1.2 \times 10^{-4} \, \text{M} \) and \( 0.24 \times 10^{-3} \, \text{M} \), respectively. The solubility product of \( \text{AB}_2 \) is:
- JEE Main - 2024
- Method of Preparation of Diazoniun Salts
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is:
- JEE Main - 2024
- Tangents and Normals
- Let \[ \vec{a} = 2\hat{i} + \alpha \hat{j} + \hat{k}, \quad \vec{b} = -\hat{i} + \hat{k}, \quad \vec{c} = \beta \hat{j} - \hat{k}, \] where \( \alpha \) and \( \beta \) are integers and \( \alpha \beta = -6 \). Let the values of the ordered pair \( (\alpha, \beta) \) for which the area of the parallelogram of diagonals \( \vec{a} + \vec{b} \) and \( \vec{b} + \vec{c} \) is \( \frac{\sqrt{21}}{2} \), be \( (\alpha_1, \beta_1) \) and \( (\alpha_2, \beta_2) \). Then \( \alpha_1^2 + \beta_1^2 - \alpha_2 \beta_2 \) is equal to:
- JEE Main - 2024
- Vectors
- The molecular formula of second homologue in the homologous series of monocarboxylic acid is
- JEE Main - 2024
- Aldehydes, Ketones and Carboxylic Acids
View More Questions