Solving Excess Charge Problem at Point 1

[jincy34]

Homework Statement

The conducting box has been given an excess negative charge. The surface density of excess electrons at the center of the top surface is 31 x 10^10 electrons/m^2. What is the electric field strength at point 1?
http://img401.imageshack.us/img401/680/knightfigure27262bn2.jpg

Homework Equations

E = linear charge density/E0
linear charge density=q/L
q=N(e)

The Attempt at a Solution

First I found the charge
q=Ne
=(31*10^10 electrons/m^2)*(1.6*10^-19)
=5*10^-8 C/m^2
E=(5*10^-8 electrons/m^2)/(8.85*10^-12)
=5604.5
Can someone please conform if am right ?

 

[rohanprabhu]

please upload your images at a service like http://imageshack.us or http://bayimg.com . We are not able to see your attachment as it is pending approval...

 

[212Phy]

This is correct. We can double check by looking at the units. Finding the charge gives us:

q=Ne=(31*10^10 electrons/m^2)*(1.6*10^-19 C/electron)
=5*10^-8 C/m^2

Than dividing by the permittivity constant, 8.85*10^-12 C^2/Nm^s gets the units into the units of electric field, N/C