Question

calculate the electric potential due to a dipole whose dipole moment is 4.6×10⁻³⁰ c·m at a point r = 2.4×10⁻⁹ m away. suppose that r≫ℓ, where ℓ is the distance between the charges in the dipole. if this point is along the axis of the dipole nearer the positive charge. express your answer to two significant figures and include the appropriate units. v = value units. part b. if this point is 60° above the axis but nearer the positive charge. express your answer to two significant figures and include the appropriate units. v = value units.

Explanation:

Step1: Recall electric - potential formula for dipole

The electric potential due to an electric dipole at a point with position vector $\vec{r}$ is given by $V=\frac{1}{4\pi\epsilon_0}\frac{\vec{p}\cdot\vec{r}}{r^{3}}$, where $\vec{p}$ is the dipole moment and $\epsilon_0 = 8.85\times 10^{-12}\ C^{2}/N\cdot m^{2}$, and $\frac{1}{4\pi\epsilon_0}=9\times 10^{9}\ N\cdot m^{2}/C^{2}$. When the point is along the axis of the dipole, $\vec{p}\cdot\vec{r}=p\times r$ (assuming the direction of $\vec{p}$ and $\vec{r}$ is the same).

Step2: Calculate potential for part A

Given $p = 4.6\times 10^{-30}\ C\cdot m$ and $r = 2.4\times 10^{-9}\ m$. Using $V=\frac{1}{4\pi\epsilon_0}\frac{p\times r}{r^{3}}=\frac{1}{4\pi\epsilon_0}\frac{p}{r^{2}}$.
Substitute the values: $V=(9\times 10^{9}\ N\cdot m^{2}/C^{2})\frac{4.6\times 10^{-30}\ C\cdot m}{(2.4\times 10^{-9}\ m)^{2}}$.
First, calculate the denominator $(2.4\times 10^{-9}\ m)^{2}=5.76\times 10^{-18}\ m^{2}$.
Then, $V=(9\times 10^{9})\frac{4.6\times 10^{-30}}{5.76\times 10^{-18}}\ V$.
$V=\frac{9\times4.6\times 10^{9 - 30}}{5.76\times 10^{-18}}\ V=\frac{41.4\times 10^{-21}}{5.76\times 10^{-18}}\ V$.
$V=\frac{41.4}{5.76}\times10^{-21 + 18}\ V$.
$V\approx7.2\times 10^{-3}\ V$.

Step3: Calculate potential for part B

When the point is at an angle $\theta = 60^{\circ}$ with the axis of the dipole, $\vec{p}\cdot\vec{r}=p\times r\times\cos\theta$.
The potential formula is $V=\frac{1}{4\pi\epsilon_0}\frac{\vec{p}\cdot\vec{r}}{r^{3}}=\frac{1}{4\pi\epsilon_0}\frac{p\times r\times\cos\theta}{r^{3}}=\frac{1}{4\pi\epsilon_0}\frac{p\times\cos\theta}{r^{2}}$.
Substitute $p = 4.6\times 10^{-30}\ C\cdot m$, $r = 2.4\times 10^{-9}\ m$, $\cos\theta=\cos60^{\circ}=0.5$, and $\frac{1}{4\pi\epsilon_0}=9\times 10^{9}\ N\cdot m^{2}/C^{2}$.
$V=(9\times 10^{9}\ N\cdot m^{2}/C^{2})\frac{4.6\times 10^{-30}\ C\cdot m\times0.5}{(2.4\times 10^{-9}\ m)^{2}}$.
We already know $(2.4\times 10^{-9}\ m)^{2}=5.76\times 10^{-18}\ m^{2}$.
$V=(9\times 10^{9})\frac{4.6\times 10^{-30}\times0.5}{5.76\times 10^{-18}}\ V$.
$V=\frac{9\times4.6\times0.5\times 10^{9 - 30}}{5.76\times 10^{-18}}\ V=\frac{20.7\times 10^{-21}}{5.76\times 10^{-18}}\ V$.
$V=\frac{20.7}{5.76}\times10^{-21 + 18}\ V\approx3.6\times 10^{-3}\ V$.

Answer:

Part A:
$V = 7.2\times 10^{-3}\ V$
Part B:
$V = 3.6\times 10^{-3}\ V$