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

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Answer:

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