A solid sphere of radius 40.0 cm has a total positive charge of 26.6 µC uniforml

Question

A solid sphere of radius 40.0 cm has a total positive charge of 26.6 µC uniformly distributed throughout its volume. Calculate the magnitude of the electric field at the following distances.

A solid sphere of radius 40.0 cm has a total positive charge of 26.6 HC uniformly distributed throughout its volume. Calculate the magnitude of the electric field at the following distances (a) 0 cm from the center of the sphere (b) 10.0 cm from the center of the sphere kN/C (c) 40.0 cm from the center of the sphere N/C (d) 61.0 cm from the center of the sphere kN/C

Explanation / Answer

using gauss' law:

electric flux through a surface=total charge enclosed by that surface

part a:

as charge enclosed by a gaussian surface of radius=0 cm is 0 C

electric field is 0 N/C.

part b:

consider a concentric sphere of radius 10 cm

charge enclosed=ratio of volumes * total charge

=(10^3/40^3)*26.6 uC=0.415625 uC

if electric field is E,

then epsilon*E*surface area of the sphere=total charge enclosed
then epsilon*E*4*pi*0.1^2=0.415625*10^(-6)

==>E=3.7372*10^5 N/C

part c:

using same method as part b and noting charge enclosed in 40 cm sphere is 26.6 uC

epsilon*E*4*pi*0.4^2=26.6*10^(-6)

==>E=1.4948*10^6 N/C

part d:

using same method as part b and noting charge enclosed in 61 cm sphere is 26.6 uC

epsilon*E*4*pi*0.61^2=26.6*10^(-6)

==>E=6.428*10^5 N/C

note:as the charge is positive, electric field is always directed radially outwards.

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