The electric field in a region is given by E-25E0-35E0- with E0-4-0- 103 N-C- The flux of this field through a rectangular surface area 0-4 m2 parallel to Y- Z plane is in Nm2C-1

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Byju's Answer

Standard XII

Physics

Magnetic Flux

The electric ...

Question

The electric field in a region is given by $\to E = \frac{2}{5} E_{0}^i + \frac{3}{5} E_{0}^j$ with $E_{0} = 4.0 \times 10^{3} N/C$ . The flux of this field through a rectangular surface area $0.4 m^{2}$ parallel to Y– Z plane is $({in Nm}^{2} C^{- 1})$

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Solution

Since rectangular surface area is parallel to Y-Z plane, so flux of the $j$ component of the field through it will be zero.

From definition of electric flux:
$ϕ = \oint \to E . - \to d A$
$ϕ = \frac{2}{5} E_{0} \times (0.4)$
$ϕ = \frac{2}{5} \times 4 \times 10^{3} \times 0.4$
$ϕ = 640 {Nm}^{2} C^{- 1}$

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The electric field in a region is given by →E=25E0^i+35E0^j with E0=4.0×103 N/C. The flux of this field through a rectangular surface area 0.4 m2 parallel to Y– Z plane is (in Nm2C−1)

Since rectangular surface area is parallel to Y-Z plane, so flux of the j component of the field through it will be zero. From definition of electric flux: ϕ=∮→E.−→dA ϕ=25E0×(0.4) ϕ=25×4×103×0.4 ϕ=640 Nm2C−1

The electric field in a region is given by $\to E = \frac{2}{5} E_{0}^i + \frac{3}{5} E_{0}^j$ with $E_{0} = 4.0 \times 10^{3} N/C$ . The flux of this field through a rectangular surface area $0.4 m^{2}$ parallel to Y– Z plane is $({in Nm}^{2} C^{- 1})$