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Consider a uniform electric field
$$
\mathbf{E}=3 \times 10^3 \hat{\mathrm{i}} \mathrm{N} / \mathrm{C} \text {. }
$$
(a) What is the flux of this field through a square of $10 \mathrm{~cm}$ on a side whose plane is parallel to the $y z$-plane?
(b) What is the flux through the same square if the normal to its plane makes a $60^{\circ}$ angle with the $x$-axis?
$$
\mathbf{E}=3 \times 10^3 \hat{\mathrm{i}} \mathrm{N} / \mathrm{C} \text {. }
$$
(a) What is the flux of this field through a square of $10 \mathrm{~cm}$ on a side whose plane is parallel to the $y z$-plane?
(b) What is the flux through the same square if the normal to its plane makes a $60^{\circ}$ angle with the $x$-axis?
Solution:
1609 Upvotes
Verified Answer
Given, $\overrightarrow{\mathrm{E}}=3 \times 10^3 \hat{\mathrm{i}} \mathrm{NC}^{-1}$,
$\mathrm{A}=10 \times 10=100 \mathrm{~cm}^2=10^{-2} \mathrm{~m}^2$
Since surface lies in $\mathrm{Y}-\mathrm{Z}$ plane, normal to the surface is along $\mathrm{X}$-axis, $\overrightarrow{\mathrm{A}}=10^{-2} \hat{\mathrm{i}} \mathrm{m}^2$
(a) $\theta=0^{\circ}, \phi=? \quad$ (b) $\theta=60^{\circ}, \phi=?$
By relation $\phi=\overline{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}=\mathrm{EA} \cos \theta$
Substituting the values, we get,
(a) $\phi=3 \times 10^3 \times 10^{-2} \times 1=30 \mathrm{Nm}^2 / \mathrm{C}$
(b) $\phi=3 \times 10^3 \times 10^{-2} \times 0.5=15 \mathrm{Nm}^2 / \mathrm{C}$
$\mathrm{A}=10 \times 10=100 \mathrm{~cm}^2=10^{-2} \mathrm{~m}^2$
Since surface lies in $\mathrm{Y}-\mathrm{Z}$ plane, normal to the surface is along $\mathrm{X}$-axis, $\overrightarrow{\mathrm{A}}=10^{-2} \hat{\mathrm{i}} \mathrm{m}^2$
(a) $\theta=0^{\circ}, \phi=? \quad$ (b) $\theta=60^{\circ}, \phi=?$
By relation $\phi=\overline{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}=\mathrm{EA} \cos \theta$
Substituting the values, we get,
(a) $\phi=3 \times 10^3 \times 10^{-2} \times 1=30 \mathrm{Nm}^2 / \mathrm{C}$
(b) $\phi=3 \times 10^3 \times 10^{-2} \times 0.5=15 \mathrm{Nm}^2 / \mathrm{C}$
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