Question

Consider a uniform electric field $E=3 \times 10^{3} \hat{i} N / C$. What is the flux of this field through a square of side $10\, cm$ if the normal to its plane makes a $60^{\circ}$ angle with the $x$ -axis?

• A. $10 \,N C ^{-1} m ^{2}$
• B. $15 \,N C ^{-1} m ^{2}$
• C. $20 \,N C ^{-1} m ^{2}$
• D. $25\, N C ^{-1} m ^{2}$

Answer 1

Answer: (B)

Here, $\vec{E}=3 \times 10^{3} \hat{i} N / C$
i.e., field is along +ve direction of $x$ -axis.
Surface area, $S=(10 \,cm )^{2}=10^{2} cm ^{2}$
$=10^{2} \times 10^{-4} m ^{2}=10^{-2} m ^{2}$
When normal to the plane makes an angle of $60^{\circ}$ with $x$ -axis,
then $\theta=60^{\circ}$
$\phi_{E}=E S\, \cos \,\theta=3 \times 10^{3} \times 10^{-2} \cos 60^{\circ}$
$=30 \times \frac{1}{2}=15 \,N C ^{-1} m ^{2}$