RADIATION BELTS....... I think but it should be radiation belt
When the velocity is increasing the acceleration increases too
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)