- #1

harshant

- 30

- 0

You should upgrade or use an alternative browser.

- Thread starter harshant
- Start date

- #1

harshant

- 30

- 0

- #2

- 12,851

- 6,056

$$V(\rho,z)=\frac{q}{4\pi \epsilon_0}\left[\frac{1}{[\rho^2+(z-d)^2]^{1/2}}-\frac{1}{[\rho^2+(z+d)^2]^{1/2}}\right]$$

The surface charge density is given by

$$\sigma(\rho)=\epsilon_0~E_z(z=0)=- \epsilon_0 \left. \frac{\partial V(\rho,z)}{\partial z}\right|_{z=0}$$

$$\sigma(\rho)=-\frac{q}{4\pi \epsilon_0} \frac{2d}{(\rho^2+d^2)^{3/2}}$$

A ring of radius ##\rho## and width ##d\rho## bears charge ##dq=\sigma(\rho)2\pi \rho~ d\rho##. The total charge on the grounded sheet is

$$Q_{total}=\int_0^{\infty} \sigma(\rho)2\pi \rho~ d\rho$$

Do the integral and convince yourself that it is equal to ##-q##.

Share:

- Replies
- 10

- Views
- 480

- Last Post

- Replies
- 5

- Views
- 341

- Replies
- 2

- Views
- 821

- Replies
- 5

- Views
- 1K

- Last Post

- Replies
- 12

- Views
- 343

- Replies
- 9

- Views
- 1K

- Replies
- 9

- Views
- 666

- Replies
- 10

- Views
- 2K

- Last Post

- Replies
- 3

- Views
- 384

- Replies
- 1

- Views
- 340