A parallel-plate capacitor has plates of area 0.22 m2 and a separation of 2.40 c
ID: 2137764 • Letter: A
Question
A parallel-plate capacitor has plates of area 0.22 m2 and a separation of 2.40 cm. A battery charges the plates to a potential difference of 110 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
Explanation / Answer
a) C= 0A/d
C= (8.85*10^-12)(.22)/(2.4/100)
C= 8.1125*10^-11 F= 81.13 pF
b) C= k* (0A/d)
C= (4.8)*(8.1125*10^-11 F)= 3.894*10^-10 F= 389.4 pF
c) C=Q/V
Q= CV
Q= (0A/d)*V
Q= 8.1125*10^-11 * 110 = 8.924*10^-9 C
d) Q= CV
Q= (k* (0A/d))* V
Q= 3.894*10^-10 * 110= 4.2384 *10^-8 C
e) Equation of electric field w/ dielectric
E= /(k0)
= Q/A
E= Q/(Ak0)
E= 8.924*10^-9/ (.22*4.8* 8.85*10^-12)
E= 954.86 V/m
f) How to calculate
E-field(from part e)-(E_field w/out dielectric)=(E_field w/ dielectric)
E_field w/out dielectric= /0= Q/A0= (8.924*10^-9)/8.85*10^-12= 1008.36
E_field in dielectric= 954.86- 1008.36= -53.47, 53.47 V/m
g) V= Ed/k
V= (1008.36*.024)/4.8
V= 5.042 V
h) W= U
U= 1/2 *CV^2
W= (1/2)CbV^2- (1/2)CaV^2, the capacitence from part a) and part b)
W= .5(389.4*10^-12)*110^2 - .5(81.13*10^-12)*110^2= 1.87*10^-6 J
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