JEE Main 202429 JanuaryEvening Shift
- Exam
- JEE Main
- Year
- 2024
- Date
- 29 January
- Shift
- Evening Shift
- Subject
- Physics
- Chapter
- Electrostatics
JEE Main 2024 · 29 January · Evening Shift · Physics · Electrostatics
An electric field is given by ( 6 ^ i + 5 ^ j + 3 ^ k ) N C ^ - 1 ....
An electric field is given by ( 6\overset{^}{i} + 5\overset{^}{j} + 3\overset{^}{k} ) N{C}^{- 1}. The electric flux through a surface area 30\overset{^}{i}{m}^{2} lying in YZ-plane(in SI unit) is:
Options
- A
- B
- C
- D
Correct answer
C.
Step-by-step solution
Given:
\overset{\to}{E} = ( 6\overset{^}{i} + 5\overset{^}{j} + 3\overset{^}{k} ) N{C}^{- 1}
\overset{\to}{A} = 30\overset{^}{i}{m}^{2}
⇒ ϕ = ( 6\overset{^}{i} + 5\overset{^}{j} + 3\overset{^}{k} ) \cdot ( 30\overset{^}{i} ) = 180\text{SI unit}