Answer to Question #118747 in Electricity and Magnetism for shaine custodio

Question

A particle of charge 3 x 〖10〗^(-9) C 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). How much work is needed to take a particle of charge 2 x 〖10〗^(-9) C from point A to point B?

Answer 1

Potential created by a point charge $q$ at distance $r$ from it is $\varphi = k \frac{q}{r}$, where $k = \frac{1}{4 \pi \epsilon_0} = 9 \cdot 10^9 N \cdot m^2 \cdot C^{-2}$ is Coulomb's constant.

Point A is at distance $r_A = \sqrt{3^2 + 3^2} = 3 \sqrt{2}$ from charge at $(0,0)$, and point B is at distance $r_B = 2$.

Hence, $\varphi_A = \frac{9 \cdot 10^9 N m^2 C^{-2} \cdot 3 \cdot 10^{-9} C}{3 \sqrt{2} m} \approx 6.4 V$ and $\varphi_B = \frac{9 \cdot 10^9 N m^2 C^{-2} \cdot 3 \cdot 10^{-9} C}{2 m} \approx 13.5 V$.

Work needed to take charge $q = 2 \cdot 10^{-9} C$ from point A to point B is $A = q (\varphi_A - \varphi_B) \approx -1.42 \cdot 10^{-8} J$.