Question

calculate the electric potential due to a dipole whose dipole moment is 4.2×10⁻³⁰ c·m at a point r = 3.2×10⁻⁹ m away. suppose that r≫ℓ, where ℓ is the distance between the charges in the dipole. 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. 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 formula for a dipole is $V = \frac{k p\cos\theta}{r^{2}}$, where $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$ is the Coulomb's constant, $p$ 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. Given $p = 4.2\times10^{-30}\ C\cdot m$, $r = 3.2\times10^{-9}\ m$, and $\theta = 60^{\circ}$ (for Part B).

Step2: Substitute values into the formula

First, find $\cos\theta=\cos60^{\circ}=\frac{1}{2}$. Then substitute $k = 9\times10^{9}\ N\cdot m^{2}/C^{2}$, $p = 4.2\times10^{-30}\ C\cdot m$, $r = 3.2\times10^{-9}\ m$, and $\cos\theta=\frac{1}{2}$ into the formula $V=\frac{k p\cos\theta}{r^{2}}$.
\[

$$\begin{align*} V&=\frac{9\times 10^{9}\times4.2\times10^{-30}\times\frac{1}{2}}{(3.2\times10^{-9})^{2}}\\ &=\frac{9\times 10^{9}\times4.2\times10^{-30}\times\frac{1}{2}}{10.24\times10^{-18}}\\ &=\frac{9\times4.2\times10^{-21}}{2\times10.24\times10^{-18}}\\ &=\frac{37.8\times10^{-21}}{20.48\times10^{-18}}\\ &=\frac{37.8}{20.48}\times10^{-3}\\ &\approx1.8\times10^{-3}\ V \end{align*}$$

\]
For Part C, when the point is $60^{\circ}$ above the axis but nearer the negative charge, $\theta = 120^{\circ}$ and $\cos\theta=-\frac{1}{2}$.
\[

$$\begin{align*} V&=\frac{9\times 10^{9}\times4.2\times10^{-30}\times(-\frac{1}{2})}{(3.2\times10^{-9})^{2}}\\ &= - 1.8\times10^{-3}\ V \end{align*}$$

\]

Answer:

Part B: $V = 1.8\times10^{-3}\ V$
Part C: $V=- 1.8\times10^{-3}\ V$