parallel-plate capacitor has plates of area 143.0 cm 2 and a separation of 1.10

Question

parallel-plate capacitor has plates of area 143.0 cm2 and a separation of 1.10 cm. A battery charges the plates to a potential difference of 113.0 V and is then disconnected. A dielectric slab of thickness 4.95 mmand dielectric constant 4.80  is then placed symmetrically between the plates. (a) Find the capacitance before the slab is inserted.

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

(c) What is the free charge (charge on a plate) before the slab is inserted?

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

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

(f) What is the magnitude of the electric field in the dielectric?

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

(h) What is the magnitude of the external work involved in the process of inserting the slab?

Explanation / Answer

a)

Capacitance

C=eoA/d =(8.85*10-12)(143*10-4)/(0.011)

C=1.15*10-11 F or 11.5pF

b)With dielectric slab insertion ,it can be treated as 3 capacitors in series C1 ,C2 and C3

Capacitance With dielectric

C1=KeoA/d =4.8*(8.85*10-12)(143*10-4)/(4.95*10-3)

C1=1.227*10-10 F or 122.7 pF

Capacitance with out dielectric

C2=C3 =(8.85*10-12)(143*10-4)/(4.95*10-3)/(0.011-0.00495)/2

C2=C3=4.18*10-11 F or 4.18 uF

So equivlent capacitance

1/C =1/C1 + 1/C2 +1/C3 =1/122.7 +1/4.18 +1/4.18

C=2.05 pF or 2.05*10-12 F

c)

Before Slab insertion

Q=CV =11.5p*113

Q=1300 pC or 1.3 nC or 1.3*10-9 C

d)

Charge after slab is inserted

Q=CV =2.05*10-12*113

Q=231.7 pC or 2.317*10-10 C

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