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Electrostatics question

2024 · 29 Jan · Shift 2 · Q67
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  5. /2024 · 29 Jan · Shift 2 · Q67

Electrostatics question

2024 · 29 Jan · Shift 2 · Q67

JEE MainPhysicsElectrostaticsMCQ+4 / −1
An electric field is given by (6i^+5j^+3k^)N/C(6 \hat{i}+5 \hat{j}+3 \hat{k}) \mathrm{N} / \mathrm{C}(6i^+5j^​+3k^)N/C. The electric flux through a surface area 30i^ m230 \hat{i} \mathrm{~m}^230i^ m2 lying in YZ-plane (in SI unit) is :
  1. A
    60
  2. B
    90
  3. C
    180
  4. D
    150
View written solutionFree

Correct answer: C

  1. Given data

    Electric field: E⃗=6i^+5j^+3k^  N/C\vec E = 6\hat i + 5\hat j + 3\hat k \; \text{N/C}E=6i^+5j^​+3k^N/C

    Surface area vector: A⃗=30i^  m2\vec A = 30\hat i \; \text{m}^2A=30i^m2

    Since the surface lies in the YZYZYZ-plane, its area vector is along the xxx-axis, which is consistent with 30i^30\hat i30i^.

  2. Formula for electric flux

    Electric flux through a surface is: Φ=E⃗⋅A⃗\Phi = \vec E \cdot \vec AΦ=E⋅A

  3. Compute the dot product

    Φ=(6i^+5j^+3k^)⋅(30i^)\Phi = (6\hat i + 5\hat j + 3\hat k) \cdot (30\hat i)Φ=(6i^+5j^​+3k^)⋅(30i^)

    Using dot product rules: i^⋅i^=1,j^⋅i^=0,k^⋅i^=0\hat i \cdot \hat i = 1, \qquad \hat j \cdot \hat i = 0, \qquad \hat k \cdot \hat i = 0i^⋅i^=1,j^​⋅i^=0,k^⋅i^=0

    Therefore, Φ=6×30=180\Phi = 6 \times 30 = 180Φ=6×30=180

  4. Final answer

    180\boxed{180}180​

    So the correct option is C.

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