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Conisder an electric field vecE=E0hatx w...

Conisder an electric field `vecE=E_0hatx` where `E_0` is a constant .
The flux through the shaded area (as shown in the figure) due to this field is

A

(a) `2E_0a_2`

B

(b) `sqrt2E_0a^2`

C

(c) `E_0a^2`

D

(d) `(E_0a^2)/(sqrt2)`

Text Solution

Verified by Experts

The correct Answer is:
C
Given `vecE=E_0hatx`
This shows that the electric field acts along +x direction and is a constant. The area vector makes an angle of `45^@` with the electric field. Therefore the electric flux through the shaded portion whose area is
`axxsqrt2a=sqrt2a^2` is `phi=vecE.vecA=EAcostheta=E_0(sqrt2a^2)cos 45^@=E_0(sqrt(2)a^2)xx(1)/(sqrt2)=E_0a^2`
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