A parallel-plate capacitor has plates of area 0.14 m 2 and a separation of 1.2 c

Question

A parallel-plate capacitor has plates of area 0.14 m2and a separation of 1.2 cm. A battery charges the plates to a potential difference of 140 V and is then disconnected. A dielectric slab of thickness 4.9 mm and dielectric constant 5.1 is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? (b) What is the capacitance with the slab in place? What is the free charge q (c) before and (d) after the slab is inserted? What is the magnitude of the electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab?

Explanation / Answer

what is the space between the plates ;

C=eoA/d; A=Cd/eo

what is the potential difference across the plates;

V=Q/C

what is the stored energy;

U=1/2CV^2

how much external work is involved in inserting slab;

Vmax=Emaxd; Qmax=CEmaxd

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