- #1
Tufts
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This is a problem I found while studying for an exam I have next thursday, and conceptually it doesn't make any sense for me!
NOTE - This is not a dieletrics problem.
A conducting slab with a thickness d and area A is inserted into the space between the plates of a parallel-plate-capacitor with spacing s and area A. What is the value of the capacitance of the system?
Diagram of the system:
------------------------
//////////////////////////// < Conducter Slab (d wide, with area A)
------------------------
[tex]C = \frac{\epsilon_{o}A}{d}[/tex] (1)
Capacitors in parallel combination: [tex]C_{eq} = C_{1} + C_{2}[/tex] (2)
Capacitors in series combination: [tex]\frac{1}{C_{eq}}= \frac{1}{C_{1}}+\frac{1}{C_{2}}[/tex] (3)
At first sight, I immediately identified this problem as a simple series capacitor combination problem. Why? The conducter slab inbetween the plates will divide the existing capacitor into two after reaching eletrostatic equilibrium, as in the diagram:
+Q
--------------------
< First capacitor
-Q
--------------------
///////////////////////
--------------------
+Q
< Second capacitor
--------------------
-Q
Since the slab is a conductor, it acts as a bridge or a conducting wire between these two capacitors. One capacitor after the other, seems to be series for me. Knowing that, I used equation 1 (where [tex]d[/tex] = 2(s - d)) to find the individual values of the capacitors, and used it along with equation 3 to find the equivalent capacitor.
My answer: [tex]C_{eq} = \frac{\epsilon_{o}A}{4(s-d)}[/tex]
Unfortunately that wasn't correct. The real answer was [tex]\frac{\epsilon_{o}A}{s-d}[/tex] which after reverse engineering, used parallel combination instead of series. Like I said before, this doesn't make sense at all for me.
Can anyone explain this? I would appreciate any help.
NOTE - This is not a dieletrics problem.
Homework Statement
A conducting slab with a thickness d and area A is inserted into the space between the plates of a parallel-plate-capacitor with spacing s and area A. What is the value of the capacitance of the system?
Diagram of the system:
------------------------
//////////////////////////// < Conducter Slab (d wide, with area A)
------------------------
Homework Equations
[tex]C = \frac{\epsilon_{o}A}{d}[/tex] (1)
Capacitors in parallel combination: [tex]C_{eq} = C_{1} + C_{2}[/tex] (2)
Capacitors in series combination: [tex]\frac{1}{C_{eq}}= \frac{1}{C_{1}}+\frac{1}{C_{2}}[/tex] (3)
The Attempt at a Solution
At first sight, I immediately identified this problem as a simple series capacitor combination problem. Why? The conducter slab inbetween the plates will divide the existing capacitor into two after reaching eletrostatic equilibrium, as in the diagram:
+Q
--------------------
< First capacitor
-Q
--------------------
///////////////////////
--------------------
+Q
< Second capacitor
--------------------
-Q
Since the slab is a conductor, it acts as a bridge or a conducting wire between these two capacitors. One capacitor after the other, seems to be series for me. Knowing that, I used equation 1 (where [tex]d[/tex] = 2(s - d)) to find the individual values of the capacitors, and used it along with equation 3 to find the equivalent capacitor.
My answer: [tex]C_{eq} = \frac{\epsilon_{o}A}{4(s-d)}[/tex]
Unfortunately that wasn't correct. The real answer was [tex]\frac{\epsilon_{o}A}{s-d}[/tex] which after reverse engineering, used parallel combination instead of series. Like I said before, this doesn't make sense at all for me.
Can anyone explain this? I would appreciate any help.