A parallel-plate capacitor has plates of area 0.22 m2 and a separation of 2.40 c

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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