12 protons in the nucleus
Answer:
(a) 23.946 kV
(b) -0.077 J
Explanation:
(a) The electric potential is given by the following formula:
(1)
k: Coulomb's constant = 8.98*10^9 Nm^2/C^2
q1 = q2 = 1.60*10^{-6}C
r1 and r2 are the distance from the charges to the point in which electric potential is evaluated.
Firs, you calculate the distance r1 and r2 by taking into account the position of the charges

Next, you replace the values of the parameters to calculate V:

(b) The potential electric energy is given by:
![U_T=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}\\\\r_{1,2}=2.00m\\\\r_{1,3}=1.20m\\\\r_{2,3}=1.20m\\\\U_T=(8.98*10^9)[\frac{(1.6*10^{-6})^2}{2.00m}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}]J\\\\U_T=-0.077J](https://tex.z-dn.net/?f=U_T%3DU_%7B1%2C2%7D%2BU_%7B1%2C3%7D%2BU_%7B2%2C3%7D%5C%5C%5C%5CU_T%3Dk%5Cfrac%7Bq_1q_2%7D%7Br_%7B1%2C2%7D%7D%2Bk%5Cfrac%7Bq_1q_3%7D%7Br_%7B1%2C3%7D%7D%2Bk%5Cfrac%7Bq_2q_3%7D%7Br_%7B2%2C3%7D%7D%5C%5C%5C%5Cr_%7B1%2C2%7D%3D2.00m%5C%5C%5C%5Cr_%7B1%2C3%7D%3D1.20m%5C%5C%5C%5Cr_%7B2%2C3%7D%3D1.20m%5C%5C%5C%5CU_T%3D%288.98%2A10%5E9%29%5B%5Cfrac%7B%281.6%2A10%5E%7B-6%7D%29%5E2%7D%7B2.00m%7D%2B%5Cfrac%7B%281.6%2A10%5E%7B-6%7D%29%28-3.70%2A10%5E%7B-6%7D%29%7D%7B1.20%7D%2B%5Cfrac%7B%281.6%2A10%5E%7B-6%7D%29%28-3.70%2A10%5E%7B-6%7D%29%7D%7B1.20%7D%5DJ%5C%5C%5C%5CU_T%3D-0.077J)
If one m³ of that material holds 4,000 kg of it,
then 0.09 m³ holds
(0.09) x (4,000) = 360 kg of it
The force of gravity acting on 360 kg of anything
on the Earth's surface is
(mass) x (gravity)
= (360 kg) x (9.8 m/s²) = 3,528