Increment of capacitance of conductor

AI Thread Summary
The discussion focuses on calculating the change in capacitance when a conductor is inserted between the plates of a parallel plate capacitor. The derived formula for the increase in capacitance is ΔC = -Aε0t/σd(d-t), which differs from the textbook answer by a negative sign. The user is uncertain about the source of this discrepancy and seeks clarification on the integration steps and the physical reasoning behind the increase in capacitance. Additionally, the user notes that capacitance should be a positive quantity, prompting questions about the sign of the electric field in the context of the problem. Understanding these concepts is crucial for resolving the confusion regarding the calculations.
Richardbryant
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Homework Statement


A Sheet of conductor of thickness t and parallel faces of cross-sectional area >=A is inserted between the plates of the capacitor of a parallel plate conductor. Show that the capacitance increased by ΔC= ε0tA/d(d-t)

Homework Equations


σ,ε0,Δ

The Attempt at a Solution


Denote Co be the initial capacitance, C' be the final capacitance.
Formula used: C=Q/V E=σ/ε0
First,Co Q/Vo, Vo=- ∫E.ds (range from 0->d) , thus the result is -σd/ε0 , Co=-Qε0/σd
Similarly, Q/V' V= - ∫E.ds (range from 0->d-t) , thus the result is -σ(d-t)/ε0 , C'=-Qε0/σ(d-t)
By answer of the two equation ΔC= C'-Co=-Qε0/σ[1/(d-t)-1/d]=-Qε0t/σd(d-t)

after substituting Q=σA then ΔC=-Aε0t/σd(d-t)
As seen, my answer is differing from the books answer by a negative sign, can anyone tell me from which step(s) i proceed wrong, there for arrive to a wrong answer.
Secondly, would anyone offer mea physical explanation of inscribing a conductor which can increase the capacitance? ?
 
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I notice there something wrong when i post this thread, therefore i shall rewrite it starting from The attempt at a solution
Denote Co be the initial capacitance, C' be the final capacitance.
Formula used: C=Q/V E=σ/ε0
First,Co Q/Vo, Vo=- ∫E.ds (range from 0->d) , thus the result is -σd/ε0 , Co=-Qε0/σd
Similarly, Q/V' V= - ∫E.ds (range from 0->d-t) , thus the result is -σ(d-t)/ε0 , C'=-Qε0/σ(d-t)

By subtraction and substituting Q=σA , the answer a get has a negative sign differ from solution.
 
I notice there something wrong when i post this thread, therefore i shall rewrite it starting from The attempt at a solution
Denote Co be the initial capacitance, C' be the final capacitance.
Formula used: C=Q/V E=σ/ε0
First,Co Q/Vo, Vo=- ∫E.ds (range from 0->d) , thus the result is -σd/ε0 , Co=-Qε0/σd
Similarly, Q/V' V= - ∫E.ds (range from 0->d-t) , thus the result is -σ(d-t)/ε0 , C'=-Qε0/σ(d-t)

Thus, substitute Q=σA and subtract the two answer I got, the solution I obtained is - ε0ta/d(d-t)

Which differ from the textbook by a negative sign.
 
Richardbryant said:
I notice there something wrong when i post this thread, therefore i shall rewrite it starting from The attempt at a solution
Denote Co be the initial capacitance, C' be the final capacitance.
Formula used: C=Q/V E=σ/ε0
First,Co Q/Vo, Vo=- ∫E.ds (range from 0->d) , thus the result is -σd/ε0 , Co=-Qε0/σd
Similarly, Q/V' V= - ∫E.ds (range from 0->d-t) , thus the result is -σ(d-t)/ε0 , C'=-Qε0/σ(d-t)

Thus, substitute Q=σA and subtract the two answer I got, the solution I obtained is - ε0ta/d(d-t)

Which differ from the textbook by a negative sign.
The capacitance is supposed to be a positive quantity.
The electric field lines start from positive charges and end in negative ones. Assuming the upper plate is positive, and you integrate the electric field from from 0 to t, what is the sign of E?
upload_2017-8-30_23-42-35.png
 
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