Question

If we triple the distance d, by what factor is the force on the point charge due to the dipole in Figure 13.60 reduced? (Note that the factor is smaller than one if the force is reduced and larger than one if the force is increased.)

Short answer

The force is reduced by the factor $0.0370$.

Step-by-step solution

Given data 

We triple the distance as, $\mathrm{D}=3\mathrm{d}$.

Dipoles

When there are two differently charged ions, they interact and make a dipole.

The moment of the dipole can be calculated with the help of the separation distance d and the charges on the ions q as,

$\mathbf{p}\mathbf{=}\mathbf{2}\mathbf{qd}$

Force generated due to dipole

When the dipole is involved, the electric field is given by,

$\mathrm{E}=\frac{2\mathrm{kp}}{{\mathrm{d}}^3}$

Here k is a constant,$\mathrm{d}$is the distance of point charge from the dipole, and$p$is the dipole moment of the dipoles.

The force on the point charge can be expressed as,

$\mathrm{F}=\mathrm{QE}$

We can rewrite the above expression as,

$\begin{array}{c}\mathrm{F}=\mathrm{Q}\frac{2\mathrm{kp}}{{\mathrm{d}}^3}\\ \mathrm{F}=\frac{\mathrm{C}}{{\mathrm{d}}^3}\end{array}$

(1)

Here C is a constant.

If we triple the distance between the point charge and dipoles, the expression for force can be written as,

${\mathrm{F}}_{\mathrm{D}}=\frac{\mathrm{C}}{{\mathrm{D}}^3}$

Substitute the values in the above equation, and we get,

$\begin{array}{c}{\mathrm{F}}_{\mathrm{D}}=\frac{\mathrm{C}}{{\left(3\mathrm{d}\right)}^3}\\ {\mathrm{F}}_{\mathrm{D}}=\frac{\mathrm{C}}{27{\left(\mathrm{d}\right)}^3}\end{array}$

From equation 1, we can rewrite the above equation as,

$\begin{array}{c}{\mathrm{F}}_{\mathrm{D}}=\frac{\mathrm{F}}{27}\\ \frac{{\mathrm{F}}_{\mathrm{D}}}{\mathrm{F}}=0.0370\end{array}$

Hence, the force is reduced by the factor$0.0370$.