Answer to Question #209550 in Electric Circuits for John Mark

Question

A charge q1= 2.00 µC is located at the origin, and a charge q2 = -6.00 µC is located at (0, 3.00) m. (A) Find the total electric potential due to these charges at the point P, whose coordinates are (4.00, 0) m. (B) Find the change in potential energy of the system of two charges plus a charge q3= 3.00 µC as the latter charge moves from infinity to point P

Answer 1

a) The electrostatic potential is additive, so $V_1 = \dfrac{kq_1}{r_1}, \; V_2 = \dfrac{kq_2}{r_2}, V_{\text{tot}} = V_1+V_2$ .

The distance from the first charge to the third is 4 m, from the second charge to the third is$\sqrt{3^2+4^2} = 5$ m. So

$V_{\text{tot}} = k\left( \dfrac{q_1}{r_1} + \dfrac{q_2}{r_2} \right) , \\ V_{\text{tot}} =8.99\cdot10^{9}\,\mathrm{N\cdot m^2/C^2}\left( \dfrac{2.00\cdot10^{-6}\,\mathrm{C}}{4\,\mathrm{m}} + \dfrac{-6.00\cdot10^{-6}\,\mathrm{C}}{5\,\mathrm{m}} \right), \\ V_{\text{tot}} = -6293\,\mathrm{V}$

b) When the charge is at infinity, the total energy of the system is the energy of the first and second charges. When we add the third charge, we add the energy $W_3 = q_3V_{\text{tot}} = 3.00\cdot10^{-6}\,\mathrm{C}\cdot (-6293\,\mathrm{V}) = -1.9\cdot10^{-2}\,\mathrm{J}.$

The change in potential energy is $-1.9\cdot10^{-2}\,\mathrm{J}.$