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Answer 1 · 2017-08-08T14:53:18

#C = 22.3 times 10^(- 12)# #"F"#

We can use the equation #C = 4 pi epsilon_(0) R#; where #R# is the radius of the sphere.

The exact value of #epsilon_(0)# is #8.854187817... times 10^(- 12)# #"F m"^(- 1)#.

#Rightarrow C = 4 pi times 8.854187817... times 10^(- 12)# #"F m"^(- 1) times 20# #"cm"#

#Rightarrow C = 4 pi times 8.854187817... times 10^(- 12)# #"F m"^(- 1) times 0.2# #"m"#

#Rightarrow C = 4 pi times 1.770837563... times 10^(- 12)# #"F"#

#Rightarrow C = 7.083350254... times 10^(- 12) times pi# #"F"#

#Rightarrow C = 22.253001121... times 10^(- 12)# #"F"#

#therefore C approx 22.3 times 10^(- 12)# #"F"#

Therefore, the capacitance of an isolated metallic sphere of radius #20# #"cm"# is around #22.3 times 10^(- 12)# #"F"#.

Answer 2 · 2017-08-08T15:11:15

#22 pF#

Let the sphere carry a charge #Q#. Its potential #V# is given as

#V=Q/(4piepsilon_0R)#
where #R# is its radius and #epsilon_0# is permittivity of free space which is #=8.85xx10^-12Fcdotm^-1#

We know that capacitance is given as

#C=Q/V#
#=>C=4piepsilon_0R#

Inserting various values in SI units we get

#C=4pixx8.85xx10^-12xx0.2#
#C=22xx10^-12F#

Capacitance is #22 pF#