Question

A proton is located at the origin, and a second proton is located on the x-axis at x = 5.22 fm (1 fm = 10-15 m).
(a) Calculate the electric potential energy associated with this configuration.(b) An alpha particle (charge = 2e, mass = 6.64 10-27 kg) is now placed at (x, y) = (2.61, 2.61) fm. Calculate the electric potential energy associated with this configuration.

Answer 1

Answer:

(a)$U_e=4.41\times10^{-14}J$

(b)$U_e=2.94\times10^{-13} J$

Step-by-step explanation:

(a)The formula for the electric potential energy is:

$U_e=k_e\frac{q_1q_2}{r}$

Where $k_e$ is the Coulomb's constant and is equal to $9\times10^9Nm^2/C^2$, $q_1$ and $q_2$are the charges and $r$ is the distance between them. In this case $r=5.22\times10^{-15}m$,$q1=q2=1.60\times10^{-19}C$.

$U_e=k_e\frac{q_1q_2}{r}=(9\times10^9)\frac{(1.60\times10^{-19})(1.60\times10^{-19})}{5.22\times10^{-15}} \\U_e=(9\times10^9)\frac{25.6\times10^{-39}}{5.22\times10^{-15}}=(9\times10^9)(4.90\times10^{-24})\\U_e=4.41\times10^{-14}J$

(b) In systems with three charges, the electric potential energy is the sum of the contribution made by each pair of the charges, i.e.  the electric potential energy of the charges $q_1$ and $q_2$ plus the electric potential energy of the charges $q_2$ and $q_3$ plus the electric potential energy of the charges $q_3$ and $q_1$.

$U_e=k_e\frac{q_1q_2}{r_{12}}+k_e\frac{q_2q_3}{r_{23}}+k_e\frac{q_3q_1}{r_{31}}$

$q_1=q_2=1.60\times10^{-19}C$

$q3=2e=2(1.60\times10^{-19})=3.2\times10^{-19}C$

Since $2.61\times10^{-15}m$ is half of $5.22\times10^{-15}m$, the charge $q_3$ is at the same distance from $q_1$ as from  $q_2$ and is the hypotenuse of the triangle rectangle with two equal sides of $2.61\times10^{-15}m$.

$r_{23}=r_{31}=\sqrt{(2.61\times10^{-15})^2+(2.61\times10^{-15})^2} =3.69\times10^{-15}m$

$r_{12}=5.22\times10^{-15}m$

$U_e=k_e\frac{q_1q_2}{r_{12}}+k_e\frac{q_2q_3}{r_{23}}+k_e\frac{q_3q_1}{r_{31}}$

$U_e=k_e\frac{(1.60\times10^{-19})(1.60\times10^{-19})}{5.22\times10^{-15}}+k_e\frac{(1.60\times10^{-19})(3.20\times10^{-19})}{3.69\times10^{-15}}+k_e\frac{(3.20\times10^{-19})(1.60\times10^{-19})}{3.69\times10^{-15}}$

$U_e=k_e\frac{2.56\times10^{-38}}{5.22\times10^{-15}}+k_e\frac{5.12\times10^{-38}}{3.69\times10^{-15}}+k_e\frac{5.12\times10^{-38}}{3.69\times10^{-15}}$

$U_e=k_e4.90\times10^{-24}+k_e1.39\times10^{-23}+k_e1.39\times10^{-23}$

$U_e=(9\times10^9)(4.90\times10^{-24})+(9\times10^9)(1.39\times10^{-23})+(9\times10^9)(1.39\times10^{-23})$

$U_e=4.41\times10^{-14}+1.25\times10^{-13}+1.25\times10^{-13}$

$U_e=2.94\times10^{-13} J$