Question

a dipole consisting of charges +q and -q, with q = 4.5 nc, is fixed to the y - axis. +q is located at (0, 5.8 mm) and -q is located at the origin. what is the electric potential at a point θ = 74° west of the dipole axis and r = 0.90m from the center of the dipole? -0.080 v -0.10 v 0.046 v 0.19 v 0.080 v

Explanation:

Step1: Recall electric - dipole potential formula

The electric - dipole potential at a point with position vector $\vec{r}$ from the center of the dipole is given by $V=\frac{1}{4\pi\epsilon_0}\frac{\vec{p}\cdot\hat{r}}{r^{2}}$, where $\vec{p} = qd$ is the dipole moment ($q$ is the magnitude of each charge in the dipole and $d$ is the distance between the charges), $\hat{r}$ is the unit - vector in the direction of $\vec{r}$, and $\frac{1}{4\pi\epsilon_0}=9\times 10^{9}\ N\cdot m^{2}/C^{2}$. First, find the dipole moment $p = Qd$. Given $Q = 4.5\ nC=4.5\times 10^{-9}\ C$ and $d = 5.8\ mm = 5.8\times 10^{-3}\ m$, so $p=Qd=(4.5\times 10^{-9}\ C)\times(5.8\times 10^{-3}\ m)=2.61\times 10^{-11}\ C\cdot m$.

Step2: Use the formula for potential in terms of dipole moment

The formula for the electric potential due to a dipole at a distance $r$ from the center of the dipole and at an angle $\theta$ with respect to the dipole axis is $V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^{2}}$. Given $\theta = 74^{\circ}$, $r = 0.90\ m$, and $\frac{1}{4\pi\epsilon_0}=9\times 10^{9}\ N\cdot m^{2}/C^{2}$, $p = 2.61\times 10^{-11}\ C\cdot m$. Then $\cos\theta=\cos(74^{\circ})\approx0.276$.

Step3: Calculate the potential

$V=(9\times 10^{9}\ N\cdot m^{2}/C^{2})\times\frac{2.61\times 10^{-11}\ C\cdot m\times0.276}{(0.90\ m)^{2}}$.
\[

$$\begin{align*} V&=(9\times 10^{9})\times\frac{2.61\times 10^{-11}\times0.276}{0.81}\\ &=(9\times 10^{9})\times\frac{7.2036\times 10^{-12}}{0.81}\\ &=\frac{64.8324\times 10^{-3}}{0.81}\\ & = 0.080\ V \end{align*}$$

\]

Answer:

$0.080\ V$