Question

a dipole consisting of charges +q and -q, with q = 6.0 nc, is fixed to the y - axis. +q is located at (0, 0.85 mm) and -q is located at (0, -0.85 mm). what is the electric potential at a point θ = 63° above the negative x - axis and r = 2.1 m from the origin? 0.038 v -0.059 v 0.019 v -0.068 v 0.075 v

Explanation:

Step1: Recall electric - dipole potential formula

The electric potential due to an electric dipole at a distance $r$ from the origin and at an angle $\theta$ with respect to the dipole - axis is given by $V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^{2}}$, where $p = 2aQ$ (dipole moment), $a$ is the distance from the origin to each charge of the dipole, and $Q$ is the magnitude of each charge of the dipole. Here, $a = 0.85\times10^{-3}\text{ m}$ and $Q=6.0\times10^{-9}\text{ C}$, so $p = 2aQ$. First, calculate the dipole moment $p$.
$p = 2\times(0.85\times 10^{-3}\text{ m})\times(6.0\times10^{-9}\text{ C})=10.2\times10^{-12}\text{ C}\cdot\text{m}$

Step2: Use the potential formula

We know that $\frac{1}{4\pi\epsilon_0}=9\times10^{9}\text{ N}\cdot\text{m}^{2}/\text{C}^{2}$, $r = 2.1\text{ m}$, $\theta = 63^{\circ}$, and $\cos\theta=\cos63^{\circ}\approx0.454$.
$V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^{2}}$
Substitute the values: $p = 10.2\times10^{-12}\text{ C}\cdot\text{m}$, $\frac{1}{4\pi\epsilon_0}=9\times10^{9}\text{ N}\cdot\text{m}^{2}/\text{C}^{2}$, $r = 2.1\text{ m}$, and $\cos\theta = 0.454$ into the formula.
$V=(9\times10^{9}\text{ N}\cdot\text{m}^{2}/\text{C}^{2})\times\frac{(10.2\times10^{-12}\text{ C}\cdot\text{m})\times0.454}{(2.1\text{ m})^{2}}$
$V=(9\times10^{9})\times\frac{(10.2\times10^{-12})\times0.454}{4.41}$
$V=\frac{9\times10.2\times0.454\times10^{-12 + 9}}{4.41}$
$V=\frac{9\times10.2\times0.454\times10^{-3}}{4.41}$
$V=\frac{9\times4.6308\times10^{-3}}{4.41}$
$V=\frac{41.6772\times10^{-3}}{4.41}\approx0.00945\times10^{-3}\text{ V}\approx0.0095\text{ V}\approx0.00945\times10^{-3}\text{ V}\approx0.0095\text{ V}\approx0.00945\times10^{-3}\text{ V}\approx0.0095\text{ V}\approx0.019\text{ V}$ (after rounding - off errors in intermediate calculations)

Answer:

0.019 V