Consider an electric field overline-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-sqrt-2-E-0-a-2-E-0-a-2-displaystyle dfrac-E-0-a-2-sqrt-2-2E-0-a-2-

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Toppr Admin · Accepted Answer

Flux-xA0-through-xA0-area-Electric-xA0-field-xD7-Area-xA0-projected-x3D5-EAcos45o-xA0- -xA0- -xA0- -xA0-E0-x221A-2a2-xD7-1-x221A-2-xA0-xA0-xA0- -xA0- -xA0- -xA0-E0a2

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

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

$F l u x t h r o u g h a r e a = E l e c t r i c f i e l d \times A r e a p r o j e c t e d$
$ϕ = E A c o s 45^{o}$
$= E_{0} (\sqrt{2} a^{2}) \times \frac{1}{\sqrt{2}}$
$= E_{0} a^{2}$

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:√2E0a2E0a2E0a2√22E0a2

Flux through area=Electric field×Area projectedϕ=EAcos45o =E0(√2a2)×1√2 =E0a2

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

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

$F l u x t h r o u g h a r e a = E l e c t r i c f i e l d \times A r e a p r o j e c t e d$
$ϕ = E A c o s 45^{o}$
$= E_{0} (\sqrt{2} a^{2}) \times \frac{1}{\sqrt{2}}$
$= E_{0} a^{2}$