A parallel-plate capacitor has plates of area 152.0 cm 2 and a separation of 2.0
ID: 2092254 • Letter: A
Question
A parallel-plate capacitor has plates of area 152.0cm2 and a separation of 2.00cm. A battery charges the plates to a potential difference of 120.0V and is then disconnected. A dielectric slab of thickness 11.20mm and dielectric constant 5.00 is then placed symmetrically between the plates. Find the capacitance before the slab is inserted.
What is the capacitance with the slab in place?
What is the free charge (charge on a plate) before the slab is inserted?
What is the free charge after the slab is inserted?
What is the magnitude of the electric field in the space between the plates and dielectric?
What is the magnitude of the electric field in the dielectric?
With the slab in place, what is the potential difference across the plates?
What is the magnitude of the external work involved in the process of inserting the slab?
Answer what you canExplanation / Answer
capacitance before the slab is inserted :
C = 8.854*10^-12*152*10^-4/2*10^-2 = 6.729*10^-12 F
capacitance with the slab in place :
Air gap = 0.02 - 0.0112 = 8.8*10^-3 m
As slab is placed symmetically :
d = 4.4*10^-3
Hence :
C(air) =8.854*10^-12*152*10^-4/4.4*10^-3= 3.05865*10^-11 F
C(slab) = 5*8.854*10^-12*152*10^-4/0.0112 = 6.008*10^-11 F
C(net) =(2/3.05865*10^-11 + 1/6.008*10^-11)^-1 = 1.219*10^-11 F
the free charge (charge on a plate) before the slab is inserted :
Q =C*V = 120*6.729*10^-12 = 8.0748*10^-10 C
the free charge after the slab is inserted :
Q =C*V = 120*1.219*10^-11= 1.4628*10^-9 C
The magnitude of the electric field in the space between the plates and dielectric :
E =V/d =120/8.8*10^-3 = 13636.3636 N/m
the magnitude of the electric field in the dielectric
E =120/0.0112*5 = 2142.857 N/m
the slab in place, what is the potential difference across the plates
V =120 V
the magnitude of the external work involved in the process of inserting the slab
W = 0.5*120^2*(C1^2-C2^2) = 6.198*10^-21 J
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