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.

part c

if this point is 60 ° above the axis but nearer the negative charge. express your answer to two significant figures and include the appropriate units.

Explanation:

Step1: Recall electric - potential formula for dipole

The electric - potential due to an electric dipole at a distance $r$ from the dipole with dipole moment $p$ and angle $\theta$ between the position vector $\vec{r}$ and the dipole - moment vector $\vec{p}$ is given by $V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^{2}}$, where $\frac{1}{4\pi\epsilon_0}=9\times 10^{9}\ N\cdot m^{2}/C^{2}$, $p$ is the dipole moment, $r$ is the distance from the dipole, and $\theta$ is the angle between the dipole moment vector and the position vector of the point.

Step2: Identify given values

We are given that $p = 4.6\times 10^{-30}\ C\cdot m$, $r = 2.4\times 10^{-9}\ m$, and $\theta = 60^{\circ}$, so $\cos\theta=\cos60^{\circ}=\frac{1}{2}$.

Step3: Substitute values into the formula

\[

$$\begin{align*} V&=(9\times 10^{9}\ N\cdot m^{2}/C^{2})\frac{4.6\times 10^{-30}\ C\cdot m\times\cos60^{\circ}}{(2.4\times 10^{-9}\ m)^{2}}\\ &=(9\times 10^{9})\frac{4.6\times 10^{-30}\times\frac{1}{2}}{(2.4\times 10^{-9})^{2}}\\ &=(9\times 10^{9})\frac{2.3\times 10^{-30}}{5.76\times 10^{-18}}\\ &=\frac{9\times 2.3\times 10^{9 - 30}}{5.76\times 10^{-18}}\\ &=\frac{20.7\times 10^{-21}}{5.76\times 10^{-18}}\\ &=\frac{20.7}{5.76}\times10^{-21 + 18}\\ &\approx3.6\times10^{-3}\ V \end{align*}$$

\]

Answer:

$V = 3.6\times 10^{-3}\ V$