Consider an electric field where is a constant. The flux through the shaded area (as shown in the figure) due to this field is [2011]

Question

Consider an electric field $\vec{E} = E_{0} \hat{x}$ where $E_{0}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is [2011]

Smart Achievers · Accepted Answer

(3)

Given $\vec{E} = E_{0} \hat{x}$

i.e., the electric field $\vec{E}$ acts along the $+ x$ -direction and is constant.

Therefore, the electric flux through the shaded portion whose area $\vec{A} = a \times \sqrt{2} a = \sqrt{2} a^{2}$ is

$ϕ = \vec{E} \cdot \vec{A}$ $= E A \cos θ$ $= E_{0} (\sqrt{2} a^{2}) \cos 45 °$ $= E_{0} (\sqrt{2} a^{2}) \times (\frac{1}{\sqrt{2}})$ $= E_{0} a^{2}$ $(∵ angle between \vec{E} and \vec{A}, Q = 45 °)$

Solution

Q.
Consider an electric field $\vec{E} = E_{0} \hat{x}$ where $E_{0}$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is [2011]

1 $2 E_{0} a^{2}$

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

3 $E_{0} a^{2}$

4 $\frac{E_{0} a^{2}}{\sqrt{2}}$

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Solution

Q. Consider an electric field E→=E0x^ where E0 is a constant. The flux through the shaded area (as shown in the figure) due to this field is [2011]

1 2E0a2

2 2E0a2

3 E0a2

4 E0a22