Question

1. Two object accumulated a charge of

Answer 1

Answer:

The  magnitude of the electric force  between these two objects

will be: 181.274 N.

i.e.  F  $=181.274$ 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:

$F = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2}{r^2}$

   $=\:\frac{4.5\times \:\:10^{-6}\:\times \:\:2.8\:\times \:\:10^{-6}}{4\:\times \:\left(3.14\right)\:\times \:\left(8.85\times \:10^{-12}\right)\times \left(2.5\times \:\:10^{-2}\right)^2}$

   $=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:4\times \:8.85\pi \left(10^{-2}\times \:2.5\right)^2}$        ∵ $4.5\times \:10^{-6}\times \:2.8\times \:10^{-6}=10^{-12}\times \:12.6$

   $=\frac{10^{-12}\times \:12.6}{10^{-12}\times \:35.4\pi \left(10^{-2}\times \:2.5\right)^2}$    ∵ $\mathrm{Multiply\:the\:numbers:}\:4\times \:8.85=35.4$

$\mathrm{Cancel\:the\:common\:factor:}\:10^{-12}$

  $=\frac{12.6}{35.4\pi \left(10^{-2}\times \:2.5\right)^2}$

  $=\frac{12.6}{0.025^2\times \:35.4\pi }$        ∵ $\left(10^{-2}\times \:2.5\right)^2=0.025^2$

  $=\frac{12.6}{0.022125\pi }$        ∵ $35.4\pi\times 0.025^2=0.022125\pi$

   $=\frac{12.6}{0.06950 }$

F  $=181.274$ N

Therefore, the  magnitude of the electric force  between these two objects will be: 181.274 N.

i.e.  F  $=181.274$ N