الفيزياء العامة الجزء الثاني#موقع الفيزياء.كوم

Lectures in General Physics
4) The flux through the closed surface remains unchanged as the charge
inside the surface is moved to another position. All of these conclusions
are arrived at through an understanding of Gauss' Law.
Example 4.5
A solid conducting sphere of
radius a has a net charge +2Q. A
conducting spherical shell of
inner radius b and outer radius c
is concentric with the solid sphere
and has a net charge –Q as shown
in figure 4.18. Using Gauss’s law
find the electric field in the
regions labeled 1, 2, 3, 4 and find
the charge distribution on the
spherical shell.
Aa
A+ 2Q
Ab
A-Q
Ac
Figure 4.18
Solution
نلاحظ أن توزيع الشحنة على الكرتين لها تماثل كروي،‏ لذلك لتعيين المجال
مختلفة فإننا سنفرض أن سطح جاوس كروي الشكل نصف قطره
الكهربي عند مناطق
.r
Region (1) r < a
To find the E inside the solid sphere of radius a we construct a gaussian
surface of radius r < a
E = 0 since no charge inside the gaussian surface.
Region (2) a < r < b
we construct a spherical gaussian surface of radius r
q
dA in
∫ E r r
. =
ε ο
Dr. Hazem Falah Sakeek