Question

A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the center of the electrode?

Express your answer to two significant figures and include the appropriate units.

Answer 1

The electric field strength is
`4.5* 10^(4) N/C`

Solution:

As per the question:

Area of the electrode,
`A_(e) = 25* 25* 10^(- 4) m^(2) = 0.0625 m^(2)`

Charge, q = 50 nC =
`50* 10^(- 9) C[/etx]</p><p>Distance, x = 2 mm = [tex]2* 10^(- 3) m`

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

`\sigma = (q)/(A_(e)) = (50* 10^(- 9))/(0.0625) = 8* 10^(- 7)C/m^(2)`

Now, the electric field strength of the electrode is:

`\vec{E} = (\sigma)/(2\epsilon_(o))`

where

`\epsilon_(o) = 8.85* 10^(- 12) F/m`

`\vec{E} = (8* 10^(- 7))/(2* 8.85* 10^(- 12))`

`\vec{E} = 4.5* 10^(4) N/C`