A parallel-plate capacitor has plates of area 0.15 m2 and a separation of 1.00 cm. A...

A parallel-plate capacitor has plates of area 0.15 m² and a separation of 1.00 cm. A battery charges the plates to a potential difference of 100 V and is then disconnected. A dielectric slab of thickness 4 mm and dielectric constant 4.8 is then placed symmetrically between the plates.

(a) What is the capacitance before the slab is inserted?
pF

(b) What is the capacitance with the slab in place?
pF

(c) What is the free charge q before the slab is inserted?
C

(d) What is the free charge q after the slab is inserted?
C

(e) What is the magnitude of the electric field in the space between the plates and dielectric?
V/m

(f) What is the magnitude of the electric field in the dielectric itself?
V/m

(g) With the slab in place, what is the potential difference across the plates?
V

(h) How much external work is involved in the process of inserting the slab?
J

Expert Solution

genius_generous answered 2 years ago

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