Question

two charges are fixed to the x - axis. if $q_1 = 2.5\text{ nc}$ is located at $x = 0\text{ cm}$, and $q_2=-2.5\text{ nc}$ is located at $x = 5.0\text{ cm}$, find the electric potential at a point $(0,2.0\text{ m})$ along the y - axis.
$\bigcirc5.5\times10^{-3}\text{ v}$
$\bigcirc1.6\times10^{-5}\text{ v}$
$\bigcirc4.9\times10^{-3}\text{ v}$
$\bigcirc6.6\times10^{-4}\text{ v}$
$\bigcirc0\text{ v}$

Explanation:

Step1: Recall electric - potential formula

The electric potential due to a point - charge is given by $V = \frac{kq}{r}$, where $k=9\times 10^{9}\ N\cdot m^{2}/C^{2}$, $q$ is the charge, and $r$ is the distance from the charge to the point of interest.

Step2: Calculate distances

The first charge $q_1 = 2.5\ nC=2.5\times 10^{-9}\ C$ is at $x = 0\ cm$. The point of interest is at $(0,2.0\ m)$. The distance $r_1$ from $q_1$ to the point $(0,2.0\ m)$ is $r_1 = 2.0\ m$. The second charge $q_2=- 2.5\ nC=-2.5\times 10^{-9}\ C$ is at $x = 5.0\ cm = 0.05\ m$. Using the Pythagorean theorem, the distance $r_2$ from $q_2$ to the point $(0,2.0\ m)$ is $r_2=\sqrt{(0.05)^{2}+(2.0)^{2}}\ m\approx2.0\ m$ (since $(0.05)^{2}\ll(2.0)^{2}$).

Step3: Calculate individual potentials

For $q_1$, $V_1=\frac{kq_1}{r_1}=\frac{9\times 10^{9}\ N\cdot m^{2}/C^{2}\times2.5\times 10^{-9}\ C}{2.0\ m}=11.25\ V$. For $q_2$, $V_2=\frac{kq_2}{r_2}=\frac{9\times 10^{9}\ N\cdot m^{2}/C^{2}\times(- 2.5\times 10^{-9}\ C)}{2.0\ m}=-11.25\ V$.

Step4: Calculate total potential

The total electric potential at the point is $V = V_1 + V_2=11.25\ V+( - 11.25\ V)=0\ V$.

Answer:

$0\ V$