Q.A charge 6 μC is placed at the origin and another charge -5 μC is placed on the y axis at a position A (0, 6.0) m. a) Calculate the total electric potential at the point P whose coordinates are (8.0, 0) m. b) Calculate the work done to bring a proton from infinity to the point P. What is the significance of the sign of the work done?

[!TLDR] Superpose the two point-charge potentials at P to get Vp=2.25×10^3 V; work to bring a proton is W=qVp=3.6×10^-16 J, positive since Vp is net positive. [!ANSWER] (a …

a) Charge $q_1=6\,\mu C$ at origin O; distance from O to $P(8.0,0)$ is $8.0$ m. Charge $q_2=-5\,\mu C$ at $A(0,6.0)$; distance from A to P is $\sqrt{8.0^2+6.0^2}=\sqrt{64+36}=\sqrt{100}=10.0$ m. By superposition (section 8.4.3), $V_P=\dfrac{1}{4\pi\epsilon_0}\left(\dfrac{q_1}{8.0}+\dfrac{q_2}{10.0}\right)=9\times10^9\left(\dfrac{6\times10^{-6}}{8.0}-\dfrac{5\times10^{-6}}{10.0}\right)=9\times10^9(7.5\times10^{-7}-5\times10^{-7})=9\times10^9\times2.5\times10^{-7}=2250\text{ V}=2.25\times10^3$ V. b) Work to bring a proton ($q=1.6\times10^{-19}$ C) from infinity to P: $W=qV_P=1.6\times10^{-19}\times2250\approx3.6\times10^{-16}$ J. The work is POSITIVE, meaning an external agent must supply this energy to bring the proton in -- consistent with $V_P$ itself …

Superpose the two point-charge potentials (with correct signs and distances) at …

A common slip is to use the straight-line distance along the axes (8.0+6.0) instead of the correct Pythagorean distan …