The common<span> feature is that the atoms of </span>all elements<span> consist of electrons, protons, and neutrons. Hope this helps!</span>
Answer:
Explanation:
5p - 14 = 8p + 4
5p = 8p + 18 <-- Moving constants to one side; add the same number of +14 to both sides.
-3p = 18. <-- The same thing with the variable itself.
p = -6 <-- Divide both sides by negative 3.
Answer:
a) E = 0
b) 
Explanation:
The electric field for all points outside the spherical shell is given as follows;
a) 
From which we have;

E = 0/A = 0
E = 0
b) 


By Gauss theorem, we have;

Therefore, we get;

The electrical field outside the spherical shell


Therefore, we have;
