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# For any patch element on the G-cylinder’s curved surface, what is the angle θ?

We want to find an expression for the electric field magnitude E at radius r from the central axis of the rod, outside the rod. We could do that using the approach of (charge element dq, field vector , etc.). However, Gauss' law gives a much faster and easier (and prettier) approach.

The charge distribution and the field have cylindrical symmetry. To find the field at radius r, we take advantage of the cylindrical symmetry by enclosing a section of the rod with a concentric Gaussian cylinder of radius r and height h.

Figure 23.4.2

(If you want to find the field at a certain point near a charged object with symmetry, arrange for the point to be on a Gaussian surface that mimics that symmetry.) We can now apply Gauss' law to relate the charge enclosed by the cylinder and the net flux through the cylinder's surface:

Q1: For any patch element on the G-cylinder’s curved surface, what is the angle θ between the electric field vector E that pierces the patch and the patch’s area vector dA?

### 1 Answer

- Anonymous10 months agoFavorite Answer
dA^ = r^ dh dφ

E = k q r^ / r³

then

the angle θ between the electric field vector E^ and the vector dA^ is null

i forgot to add the other 4 questions but that was correct. thanks.