Answer:
The magnitude of the electric force between these two objects
will be: 181.274 N.
i.e. F [tex]=181.274[/tex] N
Step-by-step explanation:
As
Two object accumulated a charge of 4.5 μC and another a charge of 2.8 μC.
so
q₁ = 4.5 μC = 4.5 × 10⁻⁶ C
q₂ = 2.8 μC = 2.8 × 10⁻⁶ C
separated distance = d = 2.5 cm
Calculating the magnitude of the force between two charged objects using the formula:
[tex]F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}[/tex]
[tex]=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}[/tex]
[tex]=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}[/tex] ∵ [tex]4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6[/tex]
[tex]=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}[/tex] ∵ [tex]\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4[/tex]
[tex]\mathrm{Cancel\:the\:common\:factor:}\:10^{-12}[/tex]
[tex]=\frac{12.6}{35.4\pi \left(10^{-2}\times \:2.5\right)^2}[/tex]
[tex]=\frac{12.6}{0.025^2\times \:35.4\pi }[/tex] ∵ [tex]\left(10^{-2}\times \:2.5\right)^2=0.025^2[/tex]
[tex]=\frac{12.6}{0.022125\pi }[/tex] ∵ [tex]35.4\pi\times 0.025^2=0.022125\pi[/tex]
[tex]=\frac{12.6}{0.06950 }[/tex]
F [tex]=181.274[/tex] N
Therefore, the magnitude of the electric force between these two objects will be: 181.274 N.
i.e. F [tex]=181.274[/tex] N
