Question

1.20 le noyau dun atome de plutonium - 239 contient 94 protons. supposez que le noyau est une sphère avec un rayon de 6,64 fm et que la charge des protons est répartie uniformément à travers la sphère. à la surface du noyau, quelle est (a) la magnitude et (b) lorientation (vers lintérieur ou lextérieur) du champ électrique produit par les protons

Explanation:

Step1: Recall electric - field formula

The electric - field magnitude at the surface of a charged sphere is given by $E = k\frac{Q}{r^{2}}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $Q$ is the total charge of the sphere, and $r$ is the radius of the sphere. The total charge of the plutonium - 239 nucleus is $Q=94e$, where $e = 1.6\times10^{-19}\ C$.

Step2: Calculate the total charge

$Q = 94\times1.6\times10^{-19}\ C=1.504\times10^{-17}\ C$. The radius $r = 6.64\ fm=6.64\times10^{-15}\ m$.

Step3: Calculate the electric - field magnitude

Substitute $Q$ and $r$ into the electric - field formula:
\[

$$\begin{align*} E&=k\frac{Q}{r^{2}}\\ &=(9\times 10^{9}\ N\cdot m^{2}/C^{2})\times\frac{1.504\times 10^{-17}\ C}{(6.64\times 10^{-15}\ m)^{2}}\\ &=(9\times 10^{9})\times\frac{1.504\times 10^{-17}}{4.40896\times 10^{-29}}\\ &=\frac{13.536\times 10^{-8}}{4.40896\times 10^{-29}}\\ & = 3.07\times10^{21}\ N/C \end{align*}$$

\]

Step4: Determine the direction of the electric field

Since the charge of the protons is positive, the electric - field lines point away from the positive charge. So the direction of the electric field at the surface of the nucleus is outward.

Answer:

(a) $3.07\times 10^{21}\ N/C$
(b) Outward