Question

Two point charges are on the y-axis. A 4.50uC charge is located at y=1.25 cm, and a -2.24uC 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) The total electric potential produced by a two point charges is the sum of the electric potential produced by a 4.50uC charge and the electric potential produced by a -2.24uC charge:

$$V_{tot}=V_1+V_2,$$ $$V_{tot}=k(\dfrac{Q_1}{r_1}+\dfrac{Q_2}{r_2})=k(\dfrac{Q_1}{y_1}+\dfrac{Q_2}{y_2}),$$

here, $k=8.99\cdot10^9\ N\dfrac{m^2}{C^2}$ is the Coulomb's constant, $Q_1=4.5\cdot10^{-6}\ C$ is the first point charge, $Q_2=-2.24\cdot10^{-6}\ C$ is the second point charge, $r_1=y_1=1.25\cdot10^{-2}\ m$ is the distance from the first point charge to the origin, $r_2=y_2=1.8\cdot10^{-2}\ m$ is the distance from the second point charge to the origin.

Then, we can calculate the total electric potential at the origin:

$V_{tot}=8.99\cdot10^9\ N\dfrac{m^2}{C^2}\cdot(\dfrac{4.5\cdot10^{-6}\ C}{1.25\cdot10^{-2}\ m}+\dfrac{(-2.24\cdot10^{-6}\ C)}{1.8\cdot10^{-2}\ m})=2.12\cdot10^6\ V.$

(b) Since the point located at $x-$axis, we can find the distances $r_1$ and $r_2$ from the Pythagorean theorem:

$$r_1=\sqrt{x^2+y_1^2}=\sqrt{(1.5\cdot10^{-2})^2+(1.25\cdot10^{-2})^2}=0.019\ m,$$ $$r_2=\sqrt{x^2+y_2^2}=\sqrt{(1.5\cdot10^{-2})^2+(1.8\cdot10^{-2})^2}=0.023\ m.$$

Then, we can calculate the total electric potential at the point whose coordinates are (1.50 cm, 0):

$$V_{tot}=k(\dfrac{Q_1}{r_1}+\dfrac{Q_2}{r_2}),$$

$V_{tot}=8.99\cdot10^9\ N\dfrac{m^2}{C^2}\cdot(\dfrac{4.5\cdot10^{-6}\ C}{0.019\ m}+\dfrac{(-2.24\cdot10^{-6}\ C)}{0.023\ m})=1.25\cdot10^6\ V.$

Answer:

(a) $V_{tot}=2.12\cdot10^6\ V.$

(b) $V_{tot}=1.25\cdot10^6\ V.$