Answer:
and
Explanation:
See attached figure.
E due to sphere
E due to particule
(1)
according to the law of gauss and superposition Law:
; electric field due to the small sphere with r1=R/4
then: (2)
on the other hand, for the particule:
⇒
(3)
We replace (2) y (3) in (1):
--------------------
if R<x<2R AND
remember that
then:
solving:
but: R<x<2R
so :
Answer:
Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e ()
with this expression we can find the closest approach distance (r)