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Electrostatics question

2021 · 1 Sep · Shift 2 · Q47
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  5. /2021 · 1 Sep · Shift 2 · Q47

Electrostatics question

2021 · 1 Sep · Shift 2 · Q47

JEE MainPhysicsElectrostaticsMCQ+4 / −1
A cube is placed inside an electric field, E→=150y2j^\overrightarrow E = 150{y^2}\widehat jE=150y2j​. 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 : JEE Main 2021 (Online) 1st September Evening Shift Physics - Electrostatics Question 132 English
  1. A
    3.8 ×\times× 10 −-− 11 C
  2. B
    8.3 ×\times× 10 −-− 11 C
  3. C
    3.8 ×\times× 10 −-− 12 C
  4. D
    8.3 ×\times× 10 −-− 12 C
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