The electric field in a region is given by E- - 35 E0i- - 45 E0- with E0 - 2-0- 103NC-1- Find the flux of this field through a rectangular surface of area 0-2 m2 parallel to the y-z plane-

Mg T = D T = mg f = mg T = N N = mg ϕ = E⃗·n̂ ds E⃗ = 3/4 E_0i + 4/5 E_0î Rectangle in yz planeSo Area vector along ...

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Question

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

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Solution

$M g - T = D$
$T = m g$
$f = m g$
$T = N$
$N = m g$
$ϕ = \to E \cdot^n d s$
$\to E = 3 / 4 E_{0} i + 4 / 5 E_{0}^i$
Rectangle in $y z$ plane
So Area vector along $^i$
$ϕ = (3 / 5 E_{0}^i + 4 / 5 E_{0}^j) \cdot (0.2)^i$
$\frac{3}{5} \times E_{0} \times 0.2$
$= \frac{3}{5} \times 2 \times 10^{3} \times 0.2$
$= 0.6 \times 0.4 \times 10^{3}$
$= 240 N m^{2} c^{- 1} m$ .

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