Question

A $2.0\mathrm{g}$metal cube and a $4.0\mathrm{g}$metal cube are $6.0\mathrm{cm}$apart, measured between their centers, on a horizontal surface. For both, the coefficient of static friction is$0.65$. Both cubes, initially neutral, are charged at a rate of $7.0\mathrm{nC}/\mathrm{s}$. How long after charging begins does one cube begin to slide away? Which cube moves first?

Short answer

A $2.0g$metal cube begins to slide away at a charge of$71.4\mathrm{nC}$with time$10.2\mathrm{s}$.

A $4.0g$metal cube begins to slide away at a charge of$100nC$with time $14.3s.$

A $2.0g$metal cube moves first.

Step-by-step solution

Given values

$\mathrm{\mu }=0.65$

$\mathrm{g}=9.81\frac{\mathrm{m}}{{\mathrm{s}}^2}$

$\mathrm{r}=6.0\mathrm{cm}\to 0.060\mathrm{m}$

${\mathrm{m}}_1=2.0\mathrm{g}\to 0.0020\mathrm{kg}$

${\mathrm{m}}_2=4.0\mathrm{g}\to 0.0040\mathrm{kg}$

$\mathrm{K}=9\times {10}^9\frac{\mathrm{N}·{\mathrm{m}}^2}{{\mathrm{C}}^2}$

$\frac{\mathrm{\Delta Q}}{\mathrm{\Delta t}}=7.0\frac{\mathrm{nC}}{\mathrm{s}}$

Calculation for charge for 2.0 g metal cube and 4.0 g metal cube.

Formula is,

$\mathrm{F}={\mathrm{\mu m}}_1\mathrm{g}-\mathrm{K}\frac{\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}=0$

$0={\mathrm{\mu m}}_1\mathrm{g}-\mathrm{K}\frac{\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}$

The charges are all the same.

${\mathrm{\mu m}}_{1\mathrm{g}}=\mathrm{K}\frac{{\mathrm{q}}^2}{{\mathrm{r}}^2}$

Both side multiplied by ${\mathrm{r}}^2$

${\mathrm{r}}^2·{\mathrm{\mu m}}_1\mathrm{g}={\mathrm{r}}^2·\mathrm{K}\frac{{\mathrm{q}}^2}{{\mathrm{r}}^2}$

${\mathrm{r}}^2·{\mathrm{\mu m}}_1\mathrm{g}={\mathrm{Kq}}^2$

Both sides are split by $\mathrm{K}$,

$\frac{{\mathrm{r}}^2·{\mathrm{\mu m}}_1\mathrm{g}}{\mathrm{K}}=\frac{\mathrm{K}·{\mathrm{q}}^2}{\mathrm{K}}$$\therefore$The charging of $2.0g$metal cube is given by

$\mathrm{q}=\sqrt{\frac{{\mathrm{r}}^2·{\mathrm{\mu m}}_1\mathrm{g}}{\mathrm{K}}}$

$\mathrm{q}=\sqrt{\frac{(0.060\mathrm{m}{)}^2·(0.65\left)\right(0.0020\mathrm{kg})\left(9.81\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)}{9\times {10}^9\frac{\mathrm{N}·{\mathrm{m}}^2}{{\mathrm{C}}^2}}}$

$\mathrm{q}=7.14\times {10}^{-8}\mathrm{C}=71.4\mathrm{nC}$

role="math" localid="1649577997286" $\therefore$charging of 4.0 g metal cube is given by

role="math" localid="1649577867082" $$\begin{gathered} \mathrm{q}=\sqrt{\frac{{\mathrm{r}}^2·{\mathrm{\mu m}}_2\mathrm{g}}{\mathrm{K}}}=\sqrt{\frac{{(0.060)}^2(0.65)(0.0040)(9.81)}{9\times {10}^9}} \\ q=1.0\times {10}^{-7}C=100nC \end{gathered}$$

Calculation for charging rate for 2.0 g metal cube and 4.0 g metal cube.

Changing time for $2.0g$metal cube is,

$\mathrm{\Delta t}=\frac{\mathrm{q}}{\frac{\mathrm{\Delta Q}}{\mathrm{\Delta t}}}$

$\mathrm{\Delta t}=\frac{71.4\mathrm{nC}}{7.0\frac{\mathrm{nC}}{\mathrm{s}}}$

$\mathrm{\Delta t}=10.2\mathrm{s}$

Charging time for$4.0g$metal cube is,

$$\begin{gathered} \mathrm{\Delta t}=\frac{\mathrm{q}}{\frac{\mathrm{\Delta Q}}{\mathrm{\Delta t}}} \\ \mathrm{\Delta t}=\frac{100nC}{7.0nC/s} \\ \mathrm{\Delta t}=14.3s \end{gathered}$$

Which metal cube moves first.

A $2.0g$metal cube charges about $10.2s$whereas a $4.0g$metal cube charges about $14.3s.$$\therefore$$2.0g$metal cube moves first.