The electric field in a region is given by . The flux of the field through a rectangular surface parallel to x – z plane is 6.0 . The area of the surface is __________ . [2025]

Question

The electric field in a region is given by $\vec{E} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^{3} N / C$ . The flux of the field through a rectangular surface parallel to x – z plane is 6.0 ${Nm}^{2} C^{- 1}$ . The area of the surface is __________ ${cm}^{2}$ . [2025]

Smart Achievers · Accepted Answer

(15)

$ϕ = \vec{E} \cdot \vec{A} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^{3} \cdot A \hat{j}$

$6 = 4 \times 10^{3} A$

$A = 1.5 \times 10^{- 3} m^{2} = 15 {cm}^{2}$

Solution

Q.
The electric field in a region is given by $\vec{E} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^{3} N / C$ . The flux of the field through a rectangular surface parallel to x – z plane is 6.0 ${Nm}^{2} C^{- 1}$ . The area of the surface is __________ ${cm}^{2}$ . [2025]

Ans.
(15)

$ϕ = \vec{E} \cdot \vec{A} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^{3} \cdot A \hat{j}$

$6 = 4 \times 10^{3} A$

$A = 1.5 \times 10^{- 3} m^{2} = 15 {cm}^{2}$

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Solution

Q. The electric field in a region is given by E→=(2i^+4j^+6k^)×103 N/C. The flux of the field through a rectangular surface parallel to x – z plane is 6.0 Nm2C–1. The area of the surface is __________cm2 . [2025]

Ans. (15) ϕ=E→·A→=(2i^+4j^+6k^)×103·Aj^ 6=4×103A A=1.5×10–3m2=15 cm2

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Q.
The electric field in a region is given by $\vec{E} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^{3} N / C$ . The flux of the field through a rectangular surface parallel to x – z plane is 6.0 ${Nm}^{2} C^{- 1}$ . The area of the surface is __________ ${cm}^{2}$ . [2025]

Ans.
(15)

$ϕ = \vec{E} \cdot \vec{A} = (2 \hat{i} + 4 \hat{j} + 6 \hat{k}) \times 10^{3} \cdot A \hat{j}$

$6 = 4 \times 10^{3} A$

$A = 1.5 \times 10^{- 3} m^{2} = 15 {cm}^{2}$