4. For a point charge, we derived an expression from Gauss’s law for the electric field a distance r away. You have a point char
ge of +2q centered in a conductive spherical shell of inner diameter a and outer diameter b. You place +8q on the spherical shell. Find the following: a. What is the electric field for rb?
A point charge of +2q centered in a conductive spherical shell of inner diameter a and outer diameter b will induce - 2q charge on the inner surface and +2q charge on the outer surface of the shell. Since 8q charge has been added to the shell , this charge will reside on the outer surface of the shell. so total charge on the outer surface will be 10q. At a point less than a , the electric field will be due to +2q charge situated at the centre . The electric field will be as follows
E = k .2q / r² for r < a
= 8kq/ a²
electric field at a point r = a>b
total charge lying inside is +2q - 2q = 0 . So in the thickness of the shell , electric field will be zero as total charge inside is nil.
For a point at r > b total charge inside is 2q-2q+10q = 10q , so electric field at r which is lying outside the shell .
When Lead-207 is bombarded with neutrons, the atom acquires the neutron. Adding a neutron only changes the mass number of the atom. This does not involve change in the identity of the atom. So, Lead-207 can change into Lead-208 without change in its chemical properties.
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