91Ó°ÊÓ

At $\mathrm{t}=0$a particle$(\mathrm{m}=0.800\text{ g´Ç°ù}0.008\text{ g},\mathrm{q}=4.00\text{‰ӼC´Ç°ù}4×{10}^{−6}\text{C})$ is located on the x axis at, $\mathrm{x}=20.0\text{ c³¾´Ç°ù}0.2\text{″¾}$moving with a speed of $v=50.0\text{″¾/s}$in the positive y-direction

For a moving body to exhibit circular motion, the body must have centripetal force acting on it. Hence, in this case, the centripetal force is balanced by the electrostatic force. This determines the value of Q.

Formulae:

The magnitude of the electrostatic force between any two particles,$\mathrm{F}=\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$ (i)

The centripetal force acting on a body in circular motion, role="math" localid="1661866945126" $\mathrm{a}={\mathrm{v}}^2/\mathrm{r}$ (ii)

For the Coulomb force to be sufficient for circular motion at that distance (where $\mathrm{r}=0.200\mathrm{m}$) the following condition is given using equations (i) and (ii) as follows:

$\begin{array}{c}\frac{\mathrm{Qq}}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2}=−\frac{{\mathrm{mv}}^2}{\mathrm{r}}\text{(negativesignastheforceisinward)}\\ \mathrm{Q}=−\frac{4{\mathrm{Ï€Î\mu }}_0\mathrm{r}.{\mathrm{mv}}^2}{\mathrm{q}}\\ =\frac{\left(0.200\text{m}\right)(8.00×{10}^{−4}\text{kg}){\left(50.0\text{m/s}\right)}^2}{(9.00×{10}^9\text{N}â‹\dots {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}})(4×{10}^{−6}\text{C})}\\ =−1.11×{10}^{−5}\text{C}\end{array}$

Hence, the value of the charge is$−1.11×{10}^{−5}\text{ C}$