Q136 / 180 · COMEDK 2025 (25 May Shift) · Physics
The magnitude and direction of the electric field in a particular region is $(10 \boldsymbol{i}+4 \boldsymbol{j}+2 \boldsymbol{k}) \times 10^4 \mathrm{NC}^{-1}$. What s the electric flux through an area $(3 \boldsymbol{i}-\boldsymbol{k}) 10^{-2} \mathrm{~m}^2$
- A$1.4 \times 10^3 \mathrm{Nm}^2 \mathrm{C}^{-1}$
- Correct answerB$2.8 \times 10^3 \mathrm{Nm}^2 \mathrm{C}^{-1}$
- C$2.8 \mathrm{Nm}^2 \mathrm{C}^{-1}$
- D$2.8 \times 10^2 \mathrm{Nm}^2 \mathrm{C}^{-1}$
Solution
The electric flux $\Phi_E$ through a surface is given by the dot product of the electric field vector $\vec{E}$ and the area vector $\vec{A}$.
Given $\vec{E} = (10 \hat{i} + 4 \hat{j} + 2 \hat{k}) \times 10^4 \text{ NC}^{-1}$ and $\vec{A} = (3 \hat{i} - \hat{k}) \times 10^{-2} \text{ m}^2$.
The flux is calculated as $\Phi_E = \vec{E} \cdot \vec{A}$.
Substituting the vectors:
$\Phi_E = [(10 \hat{i} + 4 \hat{j} + 2 \hat{k}) \times 10^4] \cdot [(3 \hat{i} + 0 \hat{j} - 1 \hat{k}) \times 10^{-2}]$
$\Phi_E = (10^4 \times 10^{-2}) \times [(10 \hat{i} + 4 \hat{j} + 2 \hat{k}) \cdot (3 \hat{i} + 0 \hat{j} - 1 \hat{k})]$
$\Phi_E = 10^2 \times [(10 \times 3) + (4 \times 0) + (2 \times -1)]$
$\Phi_E = 100 \times [30 + 0 - 2]$
$\Phi_E = 100 \times 28 = 2800 \text{ Nm}^2 \text{C}^{-1}$
Expressing in scientific notation, $\Phi_E = 2.8 \times 10^3 \text{ Nm}^2 \text{C}^{-1}$.
Answer: $2.8 \times 10^3 \mathrm{Nm}^2 \mathrm{C}^{-1}$
Given $\vec{E} = (10 \hat{i} + 4 \hat{j} + 2 \hat{k}) \times 10^4 \text{ NC}^{-1}$ and $\vec{A} = (3 \hat{i} - \hat{k}) \times 10^{-2} \text{ m}^2$.
The flux is calculated as $\Phi_E = \vec{E} \cdot \vec{A}$.
Substituting the vectors:
$\Phi_E = [(10 \hat{i} + 4 \hat{j} + 2 \hat{k}) \times 10^4] \cdot [(3 \hat{i} + 0 \hat{j} - 1 \hat{k}) \times 10^{-2}]$
$\Phi_E = (10^4 \times 10^{-2}) \times [(10 \hat{i} + 4 \hat{j} + 2 \hat{k}) \cdot (3 \hat{i} + 0 \hat{j} - 1 \hat{k})]$
$\Phi_E = 10^2 \times [(10 \times 3) + (4 \times 0) + (2 \times -1)]$
$\Phi_E = 100 \times [30 + 0 - 2]$
$\Phi_E = 100 \times 28 = 2800 \text{ Nm}^2 \text{C}^{-1}$
Expressing in scientific notation, $\Phi_E = 2.8 \times 10^3 \text{ Nm}^2 \text{C}^{-1}$.
Answer: $2.8 \times 10^3 \mathrm{Nm}^2 \mathrm{C}^{-1}$
Hence, the correct answer is option: 'B'.