Parallel plate capacitor

  • Thread starter Abdul.119
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  • #1
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Homework Statement


Consider a parallel plate capacitor of area A and separation d. The plates are isloated. One has charge +Q and the other -Q. An isolated conducting plate of area A and thickness t is inserted between the plates as shown.
i71ls1.jpg

a. Find the E fields between the conductors

b. Find the potentials between the conductors

c. Using the above, find the capacitance of this new arrangement and compare with C before inserting the conducting plate.

d. Calaculate the energy density
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of the charged capacitors before and after inserting the conducting plate

Homework Equations


σ = Q / A

The Attempt at a Solution


a. From Gauss's law, the E field for parallel plates E = σ / ε = Q / Aε , for the plate that is negatively charged, E = - Q / Aε

b. From V(x) = -∫E dx (from 0 to x), The potential is V = E * d = Qd / Aε , but we want to find the potential to the slab in the middle, so substitute d with d/2 + t/2, then V = E * (d+t)/2 = (Q(d+t)/2) / Aε = Q(d+t) / 2Aε

c. The capacitance C = Q/V
C = Q / (Q(d+t) / 2Aε) = 2Aε / (d+t)
Without the slab in the middle, C = 2Aε / d

I hope my steps are correct so far, and I don't know how to find the energy density here, I believe the energy density of a field is 1/2 * ε E^2 , but not sure how the slab would affect it
 

Answers and Replies

  • #2
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There are some important requirements to recall for electrostatics in conductors. For a conducting block there will be no electric field inside the conductor. The potential inside the block is the same everywhere, constant or zero difference. All the charge will be distributed on the surface, and the electric field lines at the surface will be perpendicular to it.
Have you drawn a sketch of the field lines?
 

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