Answer to Question #161729 in Physics for Mondy

Question

A particle of charge 3.00nC is located at point (0,0) in a certain coordinate system. Assuming all lengths are in meters, calculate the potential at the following points: A (3,3) and B (0,2)

Answer 1

The potential from a point charge at a distance $r$ from it is given as follows:

$$\varphi = k\dfrac{q}{r}$$

where $q = 3\times 10^{-9}C$ is the charge, and $k = 9\times 10^9N\cdot m^2/C^2$ is the constant.

In Cartesian coordinates the distance from the origin to some point is:

$$r = \sqrt{x^2 + y^2}$$

where $x,y$ are coordinates of this point. Thus, case of point $A (3,3)$ have:

$$\varphi_A = k\dfrac{q}{\sqrt{x^2 + y^2}} = \dfrac{9\times 10^9\cdot 3\times 10^{-9}}{\sqrt{3^2 + 3^2}} =4.5\sqrt{2} \approx 6.36V\\$$

Similarly, for point $B(0,2)$ obtain:

$$\varphi_B = k\dfrac{q}{\sqrt{x^2 + y^2}} = \dfrac{9\times 10^9\cdot 3\times 10^{-9}}{\sqrt{0^2 + 2^2}} =13.5V\\$$

Answer. $\varphi_A = 4.5\sqrt{2} \approx 6.36V, \varphi_B = 13.5V$ .