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JEE Main 20211 SeptemberEvening Shift
Exam
JEE Main
Year
2021
Date
1 September
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2021 · 1 September · Evening Shift · Physics · Electrostatics

A cube is placed inside an electric field, → E = 150 y ^ 2 ^ j The...

A cube is placed inside an electric field, \overset{\to}{E} = 150{y}^{2}\overset{^}{j} The side of the cube is 0.5m0 . 5m and is placed in the field as shown in the given figure. The charge inside the cube is:

A cube is placed inside an electric field, → E = 150 y ^ 2 ^ j The side of the cube is 0 . 5m and is placed in the

Options

  1. A8.3×10−11C8 . 3 \times {10}^{- 11}C
  2. B3.8×10−11C3 . 8 \times {10}^{- 11}C
  3. C3.8×10−12C3 . 8 \times {10}^{- 12}C
  4. D8.3×10−12C8 . 3 \times {10}^{- 12}C

Correct answer

A. 8.3×10−11C8 . 3 \times {10}^{- 11}C

Step-by-step solution

A cube is placed inside an electric field, → E = 150 y ^ 2 ^ j The side of the cube is 0 . 5m and is placed in the

By gauss law ∮E→⋅ds→=qinε0∮ \overset{\to}{E} \cdot \overset{\to}{ds} = \dfrac{{q}_{\text{in}}}{{ε}_{0}}

ε0(150)(0.5)2×(0.5)2=qin{ε}_{0} ( 150 ) ( 0 . 5{)}^{2} \times ( 0 . 5{)}^{2} = {q}_{in}

qin=8.85×10−12×15016=8.3×10−11C{q}_{in} = \dfrac{8 . 85 \times {10}^{- 12} \times 150}{16} = 8 . 3 \times {10}^{- 11}C

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