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Electrostatics question

2024 · 8 Apr · Shift 1 · Q82
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  5. /2024 · 8 Apr · Shift 1 · Q82

Electrostatics question

2024 · 8 Apr · Shift 1 · Q82

JEE MainPhysicsElectrostaticsNumerical+4 / −1
An electric field, E→=2i^+6j^+8k^6\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}E=6​2i^+6j^​+8k^​ passes through the surface of 4 m24 \mathrm{~m}^24 m2 area having unit vector n^=(2i^+j^+k^6)\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right)n^=(6​2i^+j^​+k^​). The electric flux for that surface is ‾Vm\underline{\hspace{2cm}}\mathrm{Vm}​Vm.
Numerical answer
View written solutionFree

Correct answer: 12

  1. Electric flux formula

    Electric flux through a flat surface is Φ=E⃗⋅A⃗=E⃗⋅(An^)=A(E⃗⋅n^).\Phi = \vec E \cdot \vec A = \vec E \cdot (A\hat n) = A(\vec E \cdot \hat n).Φ=E⋅A=E⋅(An^)=A(E⋅n^).

  2. Given data

    E⃗=2i^+6j^+8k^6\vec E = \frac{2\hat i+6\hat j+8\hat k}{\sqrt 6}E=6​2i^+6j^​+8k^​ n^=2i^+j^+k^6\hat n = \frac{2\hat i+\hat j+\hat k}{\sqrt 6}n^=6​2i^+j^​+k^​ A=4 m2A=4\,\text{m}^2A=4m2

  3. Compute the dot product

    E⃗⋅n^=(2i^+6j^+8k^6)⋅(2i^+j^+k^6)\vec E \cdot \hat n = \left(\frac{2\hat i+6\hat j+8\hat k}{\sqrt 6}\right)\cdot \left(\frac{2\hat i+\hat j+\hat k}{\sqrt 6}\right)E⋅n^=(6​2i^+6j^​+8k^​)⋅(6​2i^+j^​+k^​)

    Multiply corresponding components: E⃗⋅n^=16(2⋅2+6⋅1+8⋅1)\vec E \cdot \hat n = \frac{1}{6}(2\cdot 2 + 6\cdot 1 + 8\cdot 1)E⋅n^=61​(2⋅2+6⋅1+8⋅1) =16(4+6+8)= \frac{1}{6}(4+6+8)=61​(4+6+8) =186=3= \frac{18}{6} = 3=618​=3

  4. Now calculate flux

    Φ=A(E⃗⋅n^)=4×3=12\Phi = A(\vec E \cdot \hat n) = 4 \times 3 = 12Φ=A(E⋅n^)=4×3=12

  5. Final answer

    12 Vm\boxed{12\ \text{Vm}}12 Vm​

  6. Comparison with stored answer

    Stored correct answer = 121212

    Our derived answer also = 121212.

    Hence, the stored answer is correct.

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