Q143 / 160 · TS EAMCET 2023 (13 May Shift 1) · Physics

The flux of the electric field $\overrightarrow{\mathrm{E}}=24 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+28 \hat{\mathrm{k}} \mathrm{NC}^{-1}$ through an area of $20 \mathrm{~m}^2$ on the yz plane is

  • Correct answer
    A
    $480 \mathrm{Nm}^2 \mathrm{C}^{-1}$
  • B
    $600 \mathrm{Nm}^2 \mathrm{C}^{-1}$
  • C
    $560 \mathrm{Nm}^2 \mathrm{C}^{-1}$
  • D
    $1640 \mathrm{Nm}^2 \mathrm{C}^{-1}$

Solution

Electric field is given by $$ \overrightarrow{\mathrm{E}}=24 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+28 \hat{\mathrm{k}} $$ Area, $\vec{A}=20 \hat{i} \mathrm{~m}^2$ Electric flux, $\overrightarrow{\mathrm{D}}=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}$ $$ \begin{aligned} & =(24 \hat{\mathrm{i}}+30 \hat{\mathrm{j}}+28 \hat{\mathrm{k}})(20 \hat{\mathrm{j}}) \\ & =480 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-1} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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