A point charge $\overline{Q}$is distance r from a dipole consisting of charges$\pm q$separated by distance sT e dipole is initially oriented so that Q is in the plane bisecting the dipole. Immediately after the dipole is released, what are (a) the magnitude of the force and (b) the magnitude of the torque on the dipole? You can assume

The figure below depicts a positive charge Q located along the perpendicular bisector of a dipole.

Here, r denotes the charge Q's distance from the dipole's center, and s denotes the dipole's length. The force on the dipole's positive charge is shown in red, while the force on the dipole's negative charge is shown in blue. In the diagram, O represents the dipole's center, and d represents the distance from the charge Q to each charge in the dipole.

Determine the distance between the charge Q and each charge in the dipole:

Consider the triangle formed by the charge Q, the dipole center O, and the positive charge.

$d^2=r^2+{\left(\frac{s}{2}\right)}^2$

$=r^2+\frac{s^2}{4}$

The sine and cosine angles of the same triangle are as follows:

$\sin \theta =\frac{r}{d}$

$=\frac{r}{\sqrt{r^2+\frac{s^2}{4}}}$

$\cos \theta =\frac{(s/2)}{d}$

$=\frac{s}{2d}$

$=\frac{s}{2\sqrt{r^2+\frac{s^2}{4}}}$

The force on the positive charge is now,

$F_{+q}=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{d^2}$

Substitute for $d^2$.

$F_{+q}=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left.r^2+\frac{s^2}{4}\right)}$

This gives the horizontal component of force due to the dipole's positive charge on the charge Q as follows:

$F_{+q}\cos \theta =\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left(r^2+\frac{s^2}{4}\right)}\left(\frac{s}{2\sqrt{r^2+\frac{s^2}{4}}}\right)$

For the approximation $r\gg s$,

$r^2+\frac{s^2}{4}=r^2$

As a result, the above horizontal component of the force equation becomes

$F_{+q}\cos \theta \approx \frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left(r^2\right)}\left(\frac{s}{2\sqrt{r^2}}\right)$

$=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left(r^2\right)}\left(\frac{s}{2r}\right)$

$=\frac{1}{8\pi {\epsilon }_0}\frac{Qqs}{r^3}$

In the same way, the vertical component of force is as follows:

$\left.F_{+q}\sin \theta =\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left(r^2+\frac{s^2}{4}\right)}\frac{r}{\sqrt{r^2+\frac{s^2}{4}}}\right)$

For the approximation $r\gg s$,

$r^2+\frac{s^2}{4}=r^2$

As a result, the vertical component of the force is as follows::

$F_{+q}\sin \theta =\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left(r^2\right)}\left(\frac{r}{\sqrt{r^2}}\right)$

$=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r^2}$

The magnitude of force on the negative charge is as follows:

$F_{-q}=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{d^2}$

Substitute $r^2+\frac{s^2}{4}$for $d^2$

$F_{-q}=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left(r^2+\frac{s^2}{4}\right)}$

As a result, the horizontal component of force is as follows::

$\left.F_{-q}\cos \theta =\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left(r^2+\frac{s^2}{4}\right)}\frac{s}{2\sqrt{r^2+\frac{s^2}{4}}}\right)$

For the approximation $r\gg s$,

$r^2+\frac{s^2}{4}=r^2$

As a result, the horizontal component of force shifts as follows..

$F_{-q}\cos \theta \approx \frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left(r^2\right)}\left(\frac{s}{2\sqrt{r^2}}\right)$

$=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left(r^2\right)}\left(\frac{s}{2r}\right)$

$=\frac{1}{8\pi {\epsilon }_0}\frac{Qqs}{r^3}$

Similarly, the vertical component of force is as follows

$\left.F_{-q}\sin \theta =\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left(r^2+\frac{s^2}{4}\right)}\frac{r}{\sqrt{r^2+\frac{s^2}{4}}}\right)$

For the approximation $r\gg s$,

$r^2+\frac{s^2}{4}=r^2$

Hence, the vertical component of force changes as follows

$F_{-q}\sin \theta =\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{\left(r^2\right)}\left(\frac{r}{\sqrt{r^2}}\right)$

$=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r^2}$

Because the sine components of the two forces are equal and opposite in direction, they cancel each other out. As a result, the total force on the dipole equals the sum of the cosine components on two charges,

$F_{\text{total}}=F_{+q}\cos \theta +F_{-q}\cos \theta$

Substitute $\frac{1}{8\pi {\epsilon }_0}\frac{Qqs}{r^3}$for $F_{+q}\cos \theta$and $\frac{1}{8\pi {\epsilon }_0}\frac{Qqs}{r^3}$for $F_{-q}\cos \theta$

$F_{\text{total}}=\frac{1}{8\pi {\epsilon }_0}\frac{Qqs}{\left(r^3\right)}+\frac{1}{8\pi {\epsilon }_0}\frac{Qqs}{\left(r^3\right)}$

$=\frac{1}{4\pi {\epsilon }_n}\frac{Qqs}{r^3}$

As a result, the total force exerted by the charge on the dipole is $\frac{1}{4\pi {\epsilon }_0}\frac{Qqs}{r^3}$

On the dipole, torque is produced by two forces that are equal in magnitude but opposite in direction. In this case, the sine components of the two forces on the two charges are the two forces with this property.

Torque on a dipole equals the product of one of the forces and their perpendicular distance.

$\tau =F_{-q}\sin \theta \left(s\right)$

You can even consider the force $F_{+q}$in this equation. In fact, both have same magnitude.

Substitute $\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r^2}$for $F_{-q}\sin \theta$in the equation $\tau =F_{-q}\sin \theta \left(s\right)$and solve for $\tau$.

localid="1651413286141" $\tau =\left(\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r^2}\right)\left(s\right)$

localid="1651413289879" $=\frac{1}{4\pi {\epsilon }_0}\frac{Qqs}{r^2}$

As a result, the torque on the dipole is localid="1651413294051" $\frac{1}{4\pi {\epsilon }_0}\frac{Qqs}{r^2}$