Consider an electric field E=E0 hatx, Where E0 is a constant. The flux through the shaded area (as shown in the figure) due to this field is

Question

Consider an electric field E=E0 hatx, Where E0 is a constant. The flux through the shaded area (as shown in the figure) due to this field is (A) 2 E0 a2 (B)  √ 2 E0 a2 (C)  E0 a2 (D) (E0 a2/√2)

Tardigrade Admin · Accepted Answer

Electric flux, ∅ = E.S or ∅ = ES cos θ Here, θ is the angle between E and S. In this question θ = 45°, because S is perpendicular to surface. E = E0 S = (√ 2 a) (a) = √ 2 a2 ∴ θ =(E0) (√ 2 a2) cos 45° = E0 a2

Q. Consider an electric field $E = E_{0} \overset{x}{^},$ Where $E_{0}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is

$2 E_{0} a^{2}$

$2 E_{0} a^{2}$

$E_{0} a^{2}$

$\frac{E _{0} a ^{2}}{2}$

Electric flux, $\emptyset = E . S$ or $\emptyset = ES cos θ$
Here, $θ$ is the angle between $E$ and $S$ .
In this question $θ = 4 5^{\circ},$ because S is perpendicular to surface.
$E = E_{0}$
$S = (2 a) (a) = 2 a^{2}$
$∴ θ = (E_{0}) (2 a^{2}) cos 4 5^{\circ} = E_{0} a^{2}$

Q. Consider an electric field E=E0​x^, Where E0​ is a constant. The flux through the shaded area (as shown in the figure) due to this field is

2E0​a2

2​E0​a2

E0​a2

2​E0​a2​

Electric flux, ∅=E.S or ∅=EScosθ Here, θ is the angle between E and S. In this question θ=45∘, because S is perpendicular to surface. E=E0​ S=(2​a)(a)=2​a2 ∴θ=(E0​)(2​a2)cos45∘=E0​a2

Q. Consider an electric field $E = E_{0} \overset{x}{^},$ Where $E_{0}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is

$2 E_{0} a^{2}$

$2 E_{0} a^{2}$

$E_{0} a^{2}$

$\frac{E _{0} a ^{2}}{2}$

Electric flux, $\emptyset = E . S$ or $\emptyset = ES cos θ$
Here, $θ$ is the angle between $E$ and $S$ .
In this question $θ = 4 5^{\circ},$ because S is perpendicular to surface.
$E = E_{0}$
$S = (2 a) (a) = 2 a^{2}$
$∴ θ = (E_{0}) (2 a^{2}) cos 4 5^{\circ} = E_{0} a^{2}$