A charged paint is spread in a very thin uniform layer over the surface of a pla

Question

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 cm. giving it a charge of -49.0 mu C. Find the electric field just inside the paint layer. Express your answer with the appropriate units. Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward. Find the electric field just outside the paint layer. Express your answer with the appropriate units. Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward. Find the electric field 5.00 cm outside the surface of the paint layer. Express your answer with the appropriate units. Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward.

Explanation / Answer

A)
using gauss law, if we make an Gaussian surface just inside the paint layer. There won't be any charge enclosed by the gaussian surface
Qenclosed = 0
Also,
E*A = Qenclosed/ebsoleneo
Since Qenclosed = 0
E = 0 N/C
Answer: 0 N/C

B)
r = 18/2 cm = 9 cm = 0.09 m
A= 4*pi*r^2 = 4*pi*(0.09)^2 = 0.102 m^2
using gauss law, if we make an Gaussian surface just inside the paint layer.
Qenclosed = 49*10^-6 C
E will be directed inward since charge is negative
E*A = Qenclosed/ensoleneo
E*0.102 = 49*10^-6 /(8.854*10^-12)
E = 5.4*10^7 N/C
Answer: 5.4*10^7 N/C
C)
r = 18/2 cm + 5 cm = 9 cm + 5cm = 14 cn = 0.14 m
A= 4*pi*r^2 = 4*pi*(0.14)^2 = 0.246 m^2
using gauss law, if we make an Gaussian surface just inside the paint layer.
Qenclosed = 49*10^-6 C
E will be directed inward since charge is negative
E*A = Qenclosed/ensoleneo
E*0.246 = 49*10^-6 /(8.854*10^-12)
E = 2.25*10^7 N/C
Answer: 2.25*10^7 N/C

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