Question: Q. 3. Given a uniform electric field $\vec{E}=5 \times 10^{3} \hat{i}$ N/C. Find the flux of this field through a square of side $10 \mathrm{~cm}$ on a side whose plane is parallel to the $y-z$ plane. What would be the flux through the same square if the plane makes a $30^{\circ}$ angle with the $x$-axis?
A [Delhi I, II, III, 2014]
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Solution:
Ans. Flux of this field through a square of $10 \mathrm{~cm} \quad 1$ Flux of the square with normal making $30^{\circ}$ angle
$$ \begin{array}{rlr} \phi & =E A \cos \theta & 1 / 2 \ \phi & =5 \times 10^{3} \times 10^{-2} \cos 0^{\circ} \mathrm{NC}^{-1} \mathrm{~m}^{2} & \ \phi & =50 \mathrm{NC}^{-1} \mathrm{~m}^{2} & 1 / 2 \ \phi & =5 \times 10^{3} \times 10^{-2} \cos 60^{\circ} \mathrm{NC}^{-1} \mathrm{~m}^{2} & 1 / 2 \ & =25 \mathrm{NC}^{-1} \mathrm{~m}^{2} & 1 / 2 \end{array} $$
[CBSE Marking Scheme, 2014]
Detailed Answer :
Given :
$E=5 \times 10^{3} \hat{\imath}$ N/C along (+) positive direction of $x$-axis.
Surface area, $A=10 \mathrm{~cm} \times 10 \mathrm{~cm}$
$$ \begin{aligned} & =0.10 \mathrm{~m} \times 0.10 \mathrm{~m} \ & =10^{-2} \mathrm{~m}^{2} \end{aligned} $$
(i) In case of plane parallel to $y-z$ plane, normal to plane will be along $x$-axis, so $\phi=0^{\circ}$
Electric flux will be calculated using $\phi=|\vec{E}| A \cos \theta$
$1 / 2$
$=5 \times 10^{3} \times 10^{-2} \times \cos 0^{\circ}$
$=50 \mathrm{Nm}^{2} / \mathrm{C}$
(ii) Since plane is making an angle of $30^{\circ}$ with $x$-axis, so normal to its plane will make $60^{\circ}$ with $x$-axis, so $\theta=60^{\circ}$ Now finding Electric flux again with $\phi=|\vec{E}| \mathrm{A} \cos \theta$ $=5 \times 10^{3} \times 10^{-2} \times \cos 60^{\circ}$ $1 / 2$ $=25 \mathrm{Nm}^{2} / \mathrm{C}$
$1 / 2$
