91Ó°ÊÓ

The radius of the sphere,$R\left(t\right)=vt$

The charge of the sphere is .$Q$

The expression for the volume charge density is given as follows.

$\mathit{ÒÏ}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{V}}$

Here,$\mathit{V}$ is the volume of the charge distribution.

The expression for the volumeof the sphere is given as follows.

$\mathit{V}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}\mathit{Ï€}{\mathit{R}}^{\mathbf{3}}$

The volume charge density is given by,

$\mathrm{ÒÏ}(r,t)=\frac{Q}{V}$

Substitute$\frac{4}{3}\mathrm{Ï€}{R}^3$for $V$into above equation.

$\begin{array}{l}\mathrm{ÒÏ}(r,t)=\frac{Q}{\frac{4}{3}\mathrm{Ï€}{R}^3}\\ \mathrm{ÒÏ}(r,t)=\frac{3Q}{4\mathrm{Ï€}{R}^3}\end{array}$

Substitute$v{t}_r$for $R$into above equation.

$\mathrm{ÒÏ}(r,t)=\frac{3Q}{4\mathrm{Ï€}{\left(v{t}_r\right)}^3}$ …… (1)

For $r<R$the density of $r$is as follows.

$\begin{array}{l}{t}_r=t−\frac{r}{c}\\ t_r=\frac{tc−r}{c}\end{array}$

Substitute$\frac{tc−r}{c}$for$t_r$into equation (1).

$\begin{array}{l}\mathrm{ÒÏ}(r,t)=\frac{3Q}{4\mathrm{Ï€}{\left(v\left(\frac{tc−r}{c}\right)\right)}^3}\\ \mathrm{ÒÏ}(r,t)=\frac{3Q{c}^3}{4\mathrm{Ï€}{v}^3}\left(\frac{1}{{(tc−r)}^3}\right)\end{array}$

The given line integral,

$Q_{eff}={∫}_0^{vt}\mathrm{ÒÏ}(r,t)d\mathrm{Ï„}$

Substitute$\frac{3Q{c}^3}{4\mathrm{Ï€}{v}^3}\left(\frac{1}{{(tc−r)}^3}\right)$for$\mathrm{ÒÏ}(r,t)$into above equation.

$\begin{array}{l}{Q}_{eff}={∫}_0^{vt}\frac{3Q{c}^3}{4\mathrm{π}{v}^3}\left(\frac{1}{{(tc−r)}^3}\right)d\mathrm{τ}\\ Q_{eff}=\frac{3Q{c}^3}{v^3}{∫}_0^{vt}\left(\frac{r^2}{{(tc−r)}^3}\right)dr\end{array}$

Here, the integration over the entire sphere and$\mathrm{θ}$,$\mathrm{ϕ}$integral give the factor of $4\mathrm{π}$.

$\begin{array}{l}{Q}_{eff}=\frac{3Q{c}^3}{v^3}{∫}_0^{vt}\left(\frac{r^2}{{(tc−r)}^3}\right)dr\\ Q_{eff}=\frac{3Q{c}^3}{v^3}[\ln \left(\frac{c}{c−v}\right)+\frac{3v^2−2cv}{2{(c−v)}^2}]\end{array}$

For ,$t>0$the expression can be reduced as,

$Q_{eff}=Q(1+\frac{9v}{4c})$

The above expression is not matched with the required result.

Similarly, for$R=vt\text{ â¶Ä‰â¶Ä‰â¶Ä‰}V<<<c$

Now,

$\begin{array}{c}v{t}_r=c(t−t_r)\\ v{t}_r=ct−c{t}_r\\ t_r(c+v)=ct\\ t_r=\frac{ct}{(c+v)}\end{array}$

Calculate the expression for the radius as,

$R=v{t}_r$

Substitute $\frac{ct}{(c+v)}$for $t_r$into above equation.

$\begin{array}{l}v{t}_r=v\frac{ct}{(c+v)}\\ v{t}_r=\frac{vct}{(c+v)}\end{array}$

The integral is given by,

$Q_{eff}={∫}_0^{vt}\mathrm{ÒÏ}(r,t)d\mathrm{Ï„}$

Substitute$\frac{vct}{(c+v)}$for$v{t}_r$and $\frac{3Q{c}^3}{4\mathrm{Ï€}{v}^3}\left(\frac{1}{{(tc−r)}^3}\right)$for$\mathrm{ÒÏ}(r,t)$into above equation.

$Q_{eff}=\frac{3Q{c}^3}{v^3}{∫}_0^{\frac{vct}{c+v}}\left(\frac{r^2}{{(ct−r)}^3}\right)dr$

Here, the integration over the entire sphere and $\mathrm{θ}$, $\mathrm{ϕ}$integral give the factor of $4\mathrm{π}$.

$Q_{eff}=\frac{3Q{c}^3}{v^3}[\ln \left(\frac{c+v}{c}\right)+\frac{v^2−2cv}{2c^2}]$

Apply the condition $v<<<c$.

$Q_{eff}=(1−\frac{3v}{4c})$

Hence the expression for the required integral value for$v<<<c$ is$Q(1−\frac{3v}{4c})$ .