Question

problem 17.33 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 a if this point is along the axis of the dipole nearer the positive charge. express your answer to two significant figures and include the appropriate units. v = value units submit previous answers request answer incorrect; try again; 2 attempts remaining part b

Explanation:

Step1: Recall electric - potential formula for dipole

The electric potential due to an electric dipole at a point along the dipole axis is given by $V=\frac{1}{4\pi\epsilon_0}\frac{p}{r^{2}}$, where $p$ is the dipole moment, $r$ is the distance from the dipole center to the point, and $\frac{1}{4\pi\epsilon_0}=9\times 10^{9}\ N\cdot m^{2}/C^{2}$.

Step2: Substitute given values

We are given that $p = 4.2\times 10^{-30}\ C\cdot m$ and $r = 3.2\times 10^{-9}\ m$. Substituting these values into the formula:
\[

$$\begin{align*} V&=(9\times 10^{9}\ N\cdot m^{2}/C^{2})\times\frac{4.2\times 10^{-30}\ C\cdot m}{(3.2\times 10^{-9}\ m)^{2}}\\ &=(9\times 10^{9})\times\frac{4.2\times 10^{-30}}{10.24\times 10^{-18}}\\ &=\frac{9\times4.2\times 10^{9 - 30}}{10.24\times 10^{-18}}\\ &=\frac{37.8\times 10^{-21}}{10.24\times 10^{-18}}\\ &=\frac{37.8}{10.24}\times10^{-21 + 18}\\ & = 3.7\times10^{-3}\ V \end{align*}$$

\]

Answer:

$3.7\times 10^{-3}\ V$