gauss law applies.
volume1=4/3 PI ((.219^3)-(.146)^3)
now to get the charge contained, that volume x 4.61nC/m^3
E=kQ/.219^2
second part, all the charge is contained
volumecharge=4/3 pi ((.438^2)-(.146)^2)
charge = volumecharge*chargedensity
= volumecharge*4.61nC/m^3
E=kQ/1.40^2
A hollow sphere has a uniform volume charge density of 4.61 nC/m3. The inner radius is a = 14.6 cm and the outer radius is b = 43.8 cm.
What is the magnitude of the electric field at 21.9 cm from the center of the sphere?
What is the magnitude of the electric field at 140. cm from the center of the sphere?
2 answers
bobpursley did you intentionally square the radius values for the second volume, instead of making it to the power of 3?
" volumecharge=4/3 pi ((.438^2)-(.146)^2) "
" volumecharge=4/3 pi ((.438^2)-(.146)^2) "
2 answers