Shigun
Jan26-10, 11:49 AM
1. The problem statement, all variables and given/known data
A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find:
(a) the charge density of each face of the plate and
(b) the total charge on each face
2. Relevant equations
E = \frac{\sigma}{2\epsilon {0}}
\sigma = \frac{q}{a}
3. The attempt at a solution
I missed about a week due to unforeseen circumstances, so I'm struggling with this to catch up, but so far:
Area, converted to meters, is 0.25 m^2. Electric field, E, converted to nano units would be 80x10^-9. My understanding was that \Phi _E = 2EA so \Phi _E = 2 \times \frac{\sigma}{2\epsilon {0}} \times \frac{q}{a} which comes out to \Phi _E = 2 \times \frac{\sigma}{8.854\times(10^-12)}\times\frac{q}{0.50\times0.50} , however I'm a bit confused on this. How do I find the value of q, and am I correct in my setup such that \sigma is setup within the second fraction, over 8.854*10^-12?
Any help is greatly appreciated
A square plate of copper with 50.0-cm sides has no net charge and is placed in a region of uniform electric field of 80.0 kN/C directed perpendicularly to the plate. Find:
(a) the charge density of each face of the plate and
(b) the total charge on each face
2. Relevant equations
E = \frac{\sigma}{2\epsilon {0}}
\sigma = \frac{q}{a}
3. The attempt at a solution
I missed about a week due to unforeseen circumstances, so I'm struggling with this to catch up, but so far:
Area, converted to meters, is 0.25 m^2. Electric field, E, converted to nano units would be 80x10^-9. My understanding was that \Phi _E = 2EA so \Phi _E = 2 \times \frac{\sigma}{2\epsilon {0}} \times \frac{q}{a} which comes out to \Phi _E = 2 \times \frac{\sigma}{8.854\times(10^-12)}\times\frac{q}{0.50\times0.50} , however I'm a bit confused on this. How do I find the value of q, and am I correct in my setup such that \sigma is setup within the second fraction, over 8.854*10^-12?
Any help is greatly appreciated