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Surface charge, electric fields, and capacitance

by lookingforhlp
Tags: capacitance, charge, electric, fields, surface
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lookingforhlp
#1
Apr14-11, 09:23 PM
P: 11
1. The problem statement, all variables and given/known data
Two large conducting sheets are charged with a positive surface charge density. They stand vertically facing each other a distance d apart.

a.) suddenly, we shrink the left sheet to a square with sides equal to d, what is the field at the point along the central axis of the smaller sheet a point directly between the two plates?

b.) By shrinking both sheets to a square with sides equal to d, let each sheet have oppositely charged sufaces with two dielectrics (K1, K2). what is the relationship between E1 and E2 for a series configuration?

c.) Alternatively for b) what is the E-Field relationship for a parallel configuration when the sheet's contact area is shared equally?

d.) what is the equivalent capacitance in each case b) c)?



2. Relevant equations

C = (K*Epsilon*Area)/d

Efield at a point from a surface charge is = integral of kq/r^2 dq

dq = sigma dA

3. The attempt at a solution

I don't really know where to begin with this....I know in between two parallel plate capacitors the field is uniform and it will change with the change in one plate but the question does not explain what the original dimensions are....and when one shrinks that means the field wont be uniform....

But how do I figure out exactly what that is?
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lookingforhlp
#2
Apr14-11, 09:24 PM
P: 11
I want to take this one step at a time so I want to address A.) first


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