- #1

- 350

- 28

$$\vec E = \frac{\lambda*L}{4\pi\epsilon*r^2}$$

$$A = 2\pi*r*L$$

$$\vec E*A = \frac{\lambda *L^2}{2\epsilon*r} \neq \frac{q_{enc}}{\epsilon}$$

What am I doing wrong?

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- I
- Thread starter annamal
- Start date

- #1

- 350

- 28

$$\vec E = \frac{\lambda*L}{4\pi\epsilon*r^2}$$

$$A = 2\pi*r*L$$

$$\vec E*A = \frac{\lambda *L^2}{2\epsilon*r} \neq \frac{q_{enc}}{\epsilon}$$

What am I doing wrong?

- #2

Mentor

- 45,500

- 2,024

Don't confuse the field from a point charge with that of a line of charge. Look it up!$$\vec E = \frac{\lambda*L}{4\pi\epsilon*r^2}$$

- #3

- 22,741

- 13,676

I guess you are considering an "infinitely long" homogeneous line-charge along the ##z## axis of a cylindrical coordinate system. Now think about the symmetry of the problem, i.e., in which direction must the electric field point, and then use an arbitrary cylindrical volume, ##V##, with the axis along the line-charge distribution and evaluate carefully Gauss's Law,

$$\int_{\partial V} \mathrm{d}^3 \vec{f} \cdot \vec{E}=\frac{1}{\epsilon_0} Q_V,$$

where ##\partial V## is the boundary of the cylinder (with the surface-normal vectors pointing out of this cylinder), and ##Q_V## is the charge inside the considered cylindrical volume, ##V##.

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