Answer to Question #138237 in Electric Circuits for Julius

Question

Two point charges are on the y-axis. A 4.50C charge is located at y=1.25 cm, and a -2.24C charge is located at y=-1.80 cm. Find the total electric potential at (a) the origin and (b) the point whose coordinates are (1.50 cm, 0).

Answer 1

A) Since the potential is by definition $\phi=\frac {W} {q}$ ($q$ - electric charge), and since the energy is $W = E\times q\times d$ , where $d$ is the distance traveled by the charge, $E$ is the tension and since $E = k\times\frac q {r ^ 2}$ , where $k$ is the electric constant, $r$ is the distance from the charge to the point, then$\phi = E\times d = k\times\frac q r$ , and the total potential is $\phi =\phi_ 1+\phi _ 2 =$ $k\times( \frac {q_1} {r_1} - \frac {q_2} {r_2} )=$ $9\times 10^9 \times( \frac {4.5}{1.25\times 10^{-2}} - \frac{2.24}{1.8\times 10^{-2}})=$ $2.12\times 10^{12}$ V.

Б) $\phi=k \times( \frac {q_1} {\sqrt{r_1^2 +1.5^2}} - \frac {q_2} {\sqrt{r_2^2 +1.5^2}} )=$$9\times 10^9 \times ( \frac {4.5} {{\sqrt{1.25^2 1.5^2}}\times 10^{-2}}$ $- \frac {2.24} {\sqrt{1.8^2 +1.5^2}\times 10^{-2}} )\approx1.21\times 10^{12}$ V.