Question

what is the electric flux through an electric field of 6×10^4 N/C which passes through a square sheet of 20cm placed at an angle of 60 degree with the field? ​

Answer 1

The solution is in the attachment.

Answer 2

Given :

→ Field intensity = 6×10$^4$N/C

→ Side of square sheet = 20cm

→ Angle with field line = 60°

To Find :

→ Electric flux passing through the square sheet.

Formula :

☞ Formula of electric flux through a plane surface area S held in a uniform electric field E is given by

$\bigstar\:\underline{\boxed{\bf{\green{\phi_E=\vec{E}\:{\tiny{\bullet}}\:\vec{S}=ES\cos\theta}}}}$

• where $\theta$ is the angle which the normal to the outward drawn normal to surface area S makes with the field E.

Calculation :

▪ 100cm = 1m

▪ 20cm = 0.2m

$\mapsto\sf\:\phi_E=ES\cos\theta\\ \\ \mapsto\sf\:\phi_E=6\times 10^4\times (0.2\times 0.2)\times \cos60\degree\\ \\ \mapsto\sf\:\phi_E=0.24\times 10^4\times \dfrac{1}{2}\\ \\ \mapsto\sf\phi_E=0.12\times 10^4\\ \\ \mapsto\underline{\boxed{\bf{\red{\phi_E=1200\:NC^{-1}m^2}}}}$