Question

A circular plane sheet of radius $10cm$ is placed in a uniform electric field of $5{\times 10}^{5}N{C}^{-1}$, making an angle of ${60}^o$ with field. The electric flux through the sheet is

• A. $1.36{\times 10}^{2}N{m}^2{C}^{-1}$
• B. $1.36{\times 10}^{4}N{m}^2{C}^{-1}$
• C. $0.515{\times 10}^{2}N{m}^2{C}^{-1}$
• D. $0.515{\times 10}^{4}N{m}^2{C}^{-1}$

Answer 1

Answer: (B)

$1.36{\times 10}^{4}N{m}^2{C}^{-1}$

Here, $r=10cm=0.1m,E=5\times {10}^{5}N{C}^{-1}$

AS the angle between the pane sheet and the electric field is ${60}^o$, angle made by the normal to the plane sheet and the electric field is

$\theta ={90}^o-{60}^o={30}^o$

${\varphi }_E=ES\cos \theta$

$=E\times \pi r^2\cos \theta$

$=5\times {10}^5\times 3.14\times {\left(0.1\right)}^2\cos {30}^o$

$=1.36\times {10}^{4}N{m}^2{C}^{-1}$