Answer
a. $6.6\times10^{-3}V$
b. $4.6\times10^{-3}V$
c. $-4.6\times10^{-3}V$
Work Step by Step
The electric potential due to the dipole is given by equation 17–6b.
a. $V=\frac{kpcos \theta}{r^2}$
Referring to Figure 17-12, in this position the angle $\theta$ is $0^{\circ}$.
$$V=\frac{kpcos \theta}{r^2}$$
$$=(8.99\times10^9\frac{N\cdot m^2}{C^2})\frac{(4.2\times10^{-30}C \cdot m)(cos0^{\circ})}{(2.4\times10^{-9}m)^2}$$
$$=6.6\times10^{-3}V$$
b. $V=\frac{kpcos \theta}{r^2}$
Referring to Figure 17-12, in this position the angle $\theta$ is $45^{\circ}$.
$$V=\frac{kpcos \theta}{r^2}$$
$$=(8.99\times10^9\frac{N\cdot m^2}{C^2})\frac{(4.2\times10^{-30}C \cdot m)(cos45^{\circ})}{(2.4\times10^{-9}m)^2}$$
$$=4.6\times10^{-3}V$$
c. $V=\frac{kpcos \theta}{r^2}$
Referring to Figure 17-12, in this position the angle $\theta$ is $135^{\circ}$.
$$V=\frac{kpcos \theta}{r^2}$$
$$=(8.99\times10^9\frac{N\cdot m^2}{C^2})\frac{(4.2\times10^{-30}C \cdot m)(cos135^{\circ})}{(2.4\times10^{-9}m)^2}$$
$$=-4.6\times10^{-3}V$$