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The electric field components in the given figure are Ex = αx1/2 and Ey = Ez = 0 in which α = 800 N C−1 m−1/2. The charge within the cube is, if net flux through the cube is 1.05 mC−1 (assume a = 0.1 ma = 0.1 m)
  • a)
    9.27 × 10−12 C
  • b)
    9.27 × 1012 C
  • c)
    6.97 × 10−12 C
  • d)
    3.97 × 1012 C
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
The electric field components in the given figure areEx = αx1/2a...
According to Gauss' law, the net  total electric flux through  a closed surface is proportional to the charge enclosed by the closed surface.
Mathematically, 
Net flux passing through closed surface =ϕ = 1.05 N m2 C−1
Total enclosed charge = qenclosed = ?
Permittivity = ε0 = 8.854 × 10−12 
Substituting the values and simplifying we get
qenclosed = ϕε0  = 1.05 × 8.854 × 10−12 
= 9.27 × 10−12 C
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The electric field components in the given figure areEx = αx1/2andEy = Ez = 0 in whichα = 800NC−1m−1/2. The charge within the cube is, if net flux through the cube is1.05m2C−1(assumea = 0.1ma = 0.1m)a)9.27 × 10−12 Cb)9.27 × 1012 Cc)6.97 × 10−12 Cd)3.97 × 1012 CCorrect answer is option 'A'. Can you explain this answer?
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The electric field components in the given figure areEx = αx1/2andEy = Ez = 0 in whichα = 800NC−1m−1/2. The charge within the cube is, if net flux through the cube is1.05m2C−1(assumea = 0.1ma = 0.1m)a)9.27 × 10−12 Cb)9.27 × 1012 Cc)6.97 × 10−12 Cd)3.97 × 1012 CCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The electric field components in the given figure areEx = αx1/2andEy = Ez = 0 in whichα = 800NC−1m−1/2. The charge within the cube is, if net flux through the cube is1.05m2C−1(assumea = 0.1ma = 0.1m)a)9.27 × 10−12 Cb)9.27 × 1012 Cc)6.97 × 10−12 Cd)3.97 × 1012 CCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The electric field components in the given figure areEx = αx1/2andEy = Ez = 0 in whichα = 800NC−1m−1/2. The charge within the cube is, if net flux through the cube is1.05m2C−1(assumea = 0.1ma = 0.1m)a)9.27 × 10−12 Cb)9.27 × 1012 Cc)6.97 × 10−12 Cd)3.97 × 1012 CCorrect answer is option 'A'. Can you explain this answer?.
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