JEE Main20211 Sep 2021Evening ShiftPhysicsElectrostaticsActual
A cube is placed inside an electric field, E → = 150 y 2 j ^ The side of the cube is 0 . 5 m and is placed in the field as shown in the given figure. The charge inside the cube is:
Options
- A8 . 3 × 10 - 11 C
- B3 . 8 × 10 - 11 C
- C3 . 8 × 10 - 12 C
- D8 . 3 × 10 - 12 C
Correct answer
A. 8 . 3 × 10 - 11 C
Step-by-step solution
By gauss law ∮ E → · d s → = q in  ε 0 ε 0 ( 150 ) ( 0 . 5 ) 2 × ( 0 . 5 ) 2 = q i n q i n = 8 . 85 × 10 - 12 × 150 16 = 8 . 3 × 10 - 11 C