The electric field in a region is given by `vecE=3/5 E_0veci+4/5E_0vecj` with `E_0=2.0 xx 10^3 N C^(-1)`. Find the flux of this field through a recatngular surface of area `0.2m^2` parallel to the y-z plane.

Explanation

To find the electric flux through a rectangular surface of area \(0.2 \, \text{m}^2\) that is parallel to the \(y-z\) plane, we can follow these steps:

### Step 1: Understand the Electric Field
The electric field is given by:
\[ \vec{E} = \frac{3}{5} E_0 \hat{i} + \frac{4}{5} E_0 \hat{j} \]
where \(E_0 = 2.0 \times 10^3 \, \text{N/C}\).
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