Answer :
Answer:
(a) 23.946 kV
(b) -0.077 J
Explanation:
(a) The electric potential is given by the following formula:
[tex]V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}[/tex] (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
[tex]r_1=\sqrt{(1.00)^2+(0.670)^2}m=1.20m\\\\r_2=\sqrt{(-1.00)^2+(0.670)^2}m=1.20m[/tex]
Next, you replace the values of the parameters to calculate V:
[tex]V=(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}+(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}\\\\V=23946.66\ V=23.946\ kV[/tex]
(b) The potential electric energy is given by:
[tex]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[/tex]
Following are the calculation for the energy:
Given:
Please find the question in the attached file.
To find:
points=?
Solution:
For point a:
Calculating the distance among the modification and point P:
[tex]\to r = \sqrt{1^2 + 0.59^2}\\\\[/tex]
[tex]= \sqrt{1^2 + 0.59^2} \\\\ = \sqrt{1 + 0.3481} \\\\ = \sqrt{1.3481} \\\\ =1.61 \ m[/tex]
Calculating the potential due to the modification at point P:
[tex]\to V = \frac{2 kq}{r}[/tex]
[tex]=\frac{ 2 (8.99\times 10^9)(1.66\times 10^{-6})}{1.161\ m}\\\\ =\frac{(2 \times 8.99\times 10^3 \times 1.66)}{1.161\ m}\\\\=\frac{29.8468 \times 10^3 }{1.161\ m}\\\\=\frac{29846.8}{1.161\ m}\\\\= 25707.83\\\\ = 25.707\ kV[/tex]
Calculating the modification of the potential energy:
[tex]\to U = qV[/tex]
[tex]= (-2.87\times 10^{-6})(25.707\times 10^3)\\\\= -2.87\times 10^{-3}\times 25.707\\\\ =73.779 \times 10^{-3} \\\\= -0.0738\ J\\[/tex]
Learn more about electric potential:
brainly.com/question/9383604
