Increment of capacitance of conductor

[Richardbryant]

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'=-Q[B][B][I]ε[/I]0/[B]σ(d-t)
By answer of the two equation [B][B]ΔC=[B][B][B][B][B][B][B][B][B][B][B][B][B][B][B] C'-[B][B][B]Co=-[B][B][B]Q[B][B][I]ε[/I]0/[B]σ[1/(d-t)-1/d]=-Q[B][B][B][B][B][I]ε[/I]0t/[B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B]σd(d-t)[/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B]
after substituting Q=σA then [B][B][B][B][B][B][B][B][B][B][B]ΔC=-A[B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][B][I]ε[/I]0t[B][B][B][B][B][B][B][B][B][B][B][B][B][B][B]/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? ?[/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B]

 

[Richardbryant]

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/[B][B][B][B]σ(d-t)[/B][/B][/B][/B]
[B][B][B][B]By subtraction and [B]substituting Q=[B][B][B][B][B][B]σA , the answer a get has a negative sign differ from solution.[/B][/B][/B][/B][/B][/B][/B][/B][/B][/B][/B]

 

[Richardbryant]

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.

 

[ehild]

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?