Question

Practice Problem BUILD What must the distance be between charges of +2.25 and -1.86 for the attractive force between them to be the same as that between charges of +4.06 and -2.11 separated by a distance of 2.16 pm?

Answer 1

According to Coloumb's Law, force of attraction between two unit charges is given by the formula,

   F= k (q₁q₂) / d²

where q₁ and q₂ are the unit point charges, d is the distance between them and k = 1/(4π $\epsilon$ ₀) , here $\epsilon$ ₀ is permittivitty of free space.

Here let q₁ = +4.06 and q₂ = -2.11 and d_a = 2.16pm

So force of attraction between them F₁  would be, k (q₁q₂) /d²

Given that, the Force of attraction, F₂ between two other charges q₃  = +2.25 and q₄ = -1.86 , separated by a distance d_b  is also equal to F₁

ie F₁ = F₂

k (q₁ q₂) / d_a² = k ( q₃ q₄ ) / d_b²

since 'k' is a constant, we can cancel it from both the sides, thus the equation becomes,

(q₁ q₂) / d_a² = (q₃ q₄) / d_b²

Substituting the values of q_1, q₂, d_a, q_3, and q₄  from the given data, we have

   ( ( +4.06) (-2.11) ) / (2.16)^{2 =}  ( (+2.25) (-1.86) ) / d_b²

-8.5666 / 4.6656 = ( ( 2.25) (- 1.86) ) / d_b²

-1.836119684 = -4.185 / d_b²

      d_b² = -4.185 / (- 1.836119684)

d_b² = 2.279263

d_b =  $\sqrt{}$2.279263

   d_b = 1.509722822 pm

Therefore, the required distance between the two charges +2.25 and -1.86, so that the force of attraction between them would be equal to that between the charges +4.06 amd -2.11 separated by a distance of 2.16 pm, is equal to 1.509722822 pm.

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