The electric field in a region is given by E -3-5 E 0 j with E 0-2-0 - 103 NC - 1- Find the flux of this field through a rectangular surface of area 0-2 m 2 parallel to the y-z plane-

Given E=3/5 E_oi +4/5 E_0j E_0 =2.0× 10^3 N/C The plane is parallel to yz plane hence only 3/5E_0j passes paependicular to the plane Where a s4/5E_0j goes paral ...

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Standard XII

Physics

Introducing Flux

The electric ...

Question

The electric field in a region is given by $\to E = \frac{3}{5} E_{0} \to j$ 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

$G i v e n E = \frac{3}{5} E_{o i} + \frac{4}{5} E_{0 j}$

$E_{0} = 2.0 \times 10^{3} N / C$

The plane is parallel to yz plane hence only

$\frac{3}{5} E_{0 j}$ passes paependicular to the plane

Where a s $\frac{4}{5} E_{0 j}$ goes parallel.

Area =0.2 $m^{2}$ (given)

So, flux = $E \times a = \frac{3}{5} \times 2 \times 10^{3} \times 0.2$

$= 0.24 \times 10^{3} N m^{2} / C$

$= 240 N m^{2} / C$

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The electric field in a region is given by →E=35E0→j with E0=2.0×103NC−1. Find the flux of this field through a rectangular surface of area 0.2m2 parallel to the y-z plane.

GivenE=35Eoi+45E0j E0=2.0×103N/C The plane is parallel to yz plane hence only 35E0j passes paependicular to the plane Where a s45E0j goes parallel. Area =0.2 m2 (given) So, flux =E×a=35×2×103×0.2 =0.24×103Nm2/C =240Nm2/C

The electric field in a region is given by $\to E = \frac{3}{5} E_{0} \to j$ 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.