A parallel-plate capacitor as shown in the figure below has plate area S and distance between the plates d. Initially, a dielectric slab with relative dielectric constant ɛr and thickness t (t

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Chapter1: Units, Trigonometry. And Vectors
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   Though I understand how to calculate the capacitance with dielectic inserted, I have no idea how to deal with problems with partially-filled in dilectrics. Not to mention the work required to remove the dielectric in either battery-connected or battery-disconnected state. Could you please help me explain? 

A parallel-plate capacitor as shown in the figure below has plate area S and distance
between the plates d. Initially, a dielectric slab with relative dielectric constant ɛr and
thickness t (t <d) is placed inside the capacitor. The capacitor is charged with an electric
potential difference V between the two plates. Suppose the fringing effect can be
neglected (no electric field outside the
(a) Derive the expression of capacitance for this partially filled capacitor.
(b) Open the switch. What is the minimal work done by an external force to take out the
capacitor).
dielectric slab?
(c) Keep the switch closed. What is the minimal work done by an external force to take
out the dielectric slab?
t
Transcribed Image Text:A parallel-plate capacitor as shown in the figure below has plate area S and distance between the plates d. Initially, a dielectric slab with relative dielectric constant ɛr and thickness t (t <d) is placed inside the capacitor. The capacitor is charged with an electric potential difference V between the two plates. Suppose the fringing effect can be neglected (no electric field outside the (a) Derive the expression of capacitance for this partially filled capacitor. (b) Open the switch. What is the minimal work done by an external force to take out the capacitor). dielectric slab? (c) Keep the switch closed. What is the minimal work done by an external force to take out the dielectric slab? t
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