91影视

a)

From the above figure,

Monopole moment

$Q=3q=2q$

Dipole moment

$$\begin{gathered} p=\left[3qaz+\left(-q\right)0\right] \\ p=3qaz \end{gathered}$$

Write the expression for total potential at a distanceincluding monopole term.

$$\begin{gathered} V\left(r\right)=V_{mQnQ}+Vdipole \\ =\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{Q}{r}+\frac{1}{4蟺{蔚}_0}\frac{p\widehat{路}r}{r^2} \\ =\frac{1}{4蟺{蔚}_0}\left[\frac{2q}{r}+\frac{3qa\cos 胃}{r^2}\right] \end{gathered}$$

Here, $p\widehat{路}r=3qa\cos 胃$

Therefore, the total potential at a distance r including monopole term is $\frac{1}{4蟺{蔚}_0}\left[\frac{2q}{r}+\frac{3qa\cos 胃}{r^2}\right]$.

b)

From the above figure,

Monopole moment

$Q=2q$

Dipole moment

$$\begin{gathered} p=-qa\left(\widehat{-}z\right) \\ =qaz \end{gathered}$$

Write the expression for total potential at a distance r including monopole term.

$$\begin{gathered} V\left(r\right)=V_{mono}+Vdipole \\ =\frac{1}{4{\mathrm{蟺蔚}}_0}\frac{Q}{r}+\frac{1}{4蟺{蔚}_0}\frac{蚁\widehat{路}r}{r^2} \end{gathered}$$

Substitute $2q$for Q and $3qa\cos 胃$for $p\widehat{路}r$.

$V\left(r\right)=\frac{1}{4蟺{蔚}_0}\left[\frac{2q}{r}+\frac{3qa\cos 胃}{r^2}\right]$

Therefore, the total potential at a distance $r$including monopole term is

$\frac{1}{4蟺{蔚}_0}\left[\frac{2q}{r}+\frac{3qa\cos 胃}{r^2}\right]$.

c)

From the above figure,

Monopole moment

$Q=2q$

Dipole moment

$p=3qa\hat{y}$

Write the expression for total potential at a distance r including monopole term.

$\begin{array}{rcl}V\left(r\right)& =& V_{mono}+V_{dipole}\\ & =& \frac{1}{4蟺{蔚}_0}\frac{Q}{r}+\frac{1}{4蟺{蔚}_0}\frac{p路\hat{r}}{r^2}\\ & =& \frac{1}{4蟺{蔚}_0}\left[\frac{2q}{r}+\frac{3qa\sin 胃\sin 蠒}{r^2}\right]\\ & & \end{array}$

Therefore, the total potential at a distance r including monopole term is

$\frac{1}{4蟺{蔚}_0}\left[\frac{2q}{r}+\frac{3qa\sin 胃\sin 蠒}{r^2}\right]$

As,

$\hat{r}=\sin 胃\cos 蠒\hat{x}+\sin 胃\sin 蠒\hat{y}+\cos 胃\hat{z}$

Then,

$\hat{y}路\hat{r}=\sin 胃\sin 蠒$.