For the indicated region, use Gauss’s Law to determine the magnitude of the elec
ID: 1431516 • Letter: F
Question
For the indicated region, use Gauss’s Law to determine the magnitude of the electric field as a function of r the distance from the point charge. This requires drawing a picture, sketching the electric field, choosing a gaussian surface, calculating the charge enclosed by the surface and solving for the magnitude of the electric field.
Please answer four questions down below and give detailed calculated steps. Thank you very much! Provide as many details as possible please. I also attached all the information you will need to answer all four questions down below.
Xercises For Exercises 1-3, consider the system described in Example 1. For the indicated region, use Gauss's Law to determine the magnitude of the electric field as a function of r the distance from the point charge. This requires drawing a picture, sketching the electric field, choosing a gaussian surface, calculating the charge enclosed by the surface and solving for the magnitude of the electric field 1. The region outside of the conductor, r b 2. The region within the conducting material, a r b. You know the answer to this one is zero, but show that GausS's Law gives that result. 3. The region inside the conducting shell, r a 4. Consider the conducting cylinder in Example 2. Show that at points outside the conductor and very close to the surface, the formula E gives the strength of the electric fieldExplanation / Answer
1)
for r > b :
considering a concentric sphere of radius r (r > b) as Gaussian surface
using Gauss Law,
total flux through surface = Qinside / e0
E . A = Qinside /e0
Qinside = +Q + (-3Q) = -2Q (minus sign means field will be radially towards centre)
E ( 4 pi r^2 ) = 2Q / e0
E = Q / (2 pi e0 r^2 )
direction radially towards centre
2). a< r < b
there is charge +Q at the centre and this is a conducting shell hence inner surface of this
conducting shell will have -Q charge and -2Q charge will go on outer surface.
(charge always lies on surface in conductors)
now for gaussian sphere with radius r (a < r < b)
Qin = Q -Q = 0
hence E.A = 0
E = 0
3.
r < a
Qin = Q
E (4 pi r^2 ) = Q / e0
E = Q / (4 pi e0 r^2 )
4. this part is already done on your example sheet.