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Byju's Answer

Standard XII

Physics

Ohm's Law in Scalar Form

The electric ...

Question

The electric field in a region is given by $\to E = \frac{3}{5} E_{0}^i + \frac{4}{5} E_{0}^j$ with $E_{0} = 2.0 \times 10^{3} N C^{- 1} .$ The flux of this field through a rectangular surface of area $0.2 m^{2}$ parallel to the yz-plane is .(Take S.I. system of units everywhere)

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Solution

The area vector can be written as
$\to A =$ $0.2^i$ ( since area is parallel to yz plane)
Electric flux $ϕ = \to E . \to A$
$ϕ = [\frac{3 E_{0}}{5}^i + \frac{4 E_{0}}{5}^j] . [0.2^i]$
$ϕ = \frac{0.6 E_{0}}{5}$
$ϕ = \frac{0.6 \times 2 \times 10^{3}}{5}$
$ϕ = 240 V m$

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The electric field in a region is given by →E=35E0 ^i+45E0 ^j with E0=2.0×103NC−1. The flux of this field through a rectangular surface of area 0.2 m2 parallel to the yz-plane is .(Take S.I. system of units everywhere)

The area vector can be written as →A=0.2 ^i( since area is parallel to yz plane) Electric flux ϕ=→E.→A ϕ=[3E05^i+4E05^j].[0.2 ^i] ϕ=0.6E05 ϕ=0.6×2×1035 ϕ=240 Vm

The electric field in a region is given by $\to E = \frac{3}{5} E_{0}^i + \frac{4}{5} E_{0}^j$ with $E_{0} = 2.0 \times 10^{3} N C^{- 1} .$ The flux of this field through a rectangular surface of area $0.2 m^{2}$ parallel to the yz-plane is .(Take S.I. system of units everywhere)

The area vector can be written as
$\to A =$ $0.2^i$ ( since area is parallel to yz plane)
Electric flux $ϕ = \to E . \to A$
$ϕ = [\frac{3 E_{0}}{5}^i + \frac{4 E_{0}}{5}^j] . [0.2^i]$
$ϕ = \frac{0.6 E_{0}}{5}$
$ϕ = \frac{0.6 \times 2 \times 10^{3}}{5}$
$ϕ = 240 V m$