Answer:
321.39 m/s
[tex]1.3772213803 \times 10^{10}\ m/s^2[/tex]
Explanation:
m = Mass = [tex]9\times 10^{-14}\ kg[/tex]
[tex]q_1[/tex] = Charge = -2.50 pC
[tex]q_2[/tex] = Charge = -3.10 pC
r = Radius = 3.75 μm
[tex]\varepsilon_{0}[/tex] = Vacuum permittivity = [tex]8.85 \times 10^{-12}\ F/m[/tex]
We have the relation
[tex]m v^{2} &=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0} 2 r}[/tex]
[tex]v =\sqrt{\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}(2 r)(m)}}[/tex]
[tex]=\sqrt{\frac{\left(-2.50 \times 10^{-12}\right)\left(-3.1 \times 10^{-12}\right)}{4(3.14)\left(8.85 \times 10^{-12}\right)\left(2 \times 3.75 \times 10^{-6}\right)\left(9 \times 10^{-14}\right)}}[/tex]
[tex]=321.39\ m/s[/tex]
The Speed they need when very far away from each other to get close enough to just touch is 321.39 m/s
We have the relation
[tex]m a=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}(2 r)^{2}}[/tex]
[tex]a=\frac{q_{1} q_{2}}{4 \pi \varepsilon_{0}(2 r)^{2}(m)}[/tex]
[tex]=\frac{\left(-2.50 \times 10^{-12}\right)\left(-3.10 \times 10^{-12}\right)}{4(3.14)\left(8.85 \times 10^{-12}\right)\left(2 \times 3.75 \times 10^{-6}\right)^{2}\left(9 \times 10^{-14}\right)}[/tex]
[tex]a=1.3772213803 \times 10^{10} \mathrm{m} / \mathrm{s}^{2}[/tex]
The acceleration is [tex]1.3772213803 \times 10^{10}\ m/s^2[/tex]
