Question

There are two charges 1 micro C and 5micro C placed at certain distance. Force applied by 1 micro coulomb on 5 micro coulomb is F1 and by 5 micro C is F2. Then, find F1/F2
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Answer 1

Given :

• First charge, q₁ = 1 micro C
• Second charge, q₂ = 5 micro C
• Force applied by q₁ on q₂ = F₁
• Force applied by q₂ on q₁ = F₂

To Find :

• $\sf \dfrac{F_1}{F_2}$

Solution :

Force applied by q₁ on q₂ = F₁

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \underline{\boxed{\pink{\bf F_1 = k \dfrac{q_1 q_2}{d^2}}}}$

Force applied by q₂ on q₁ = F₂

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \underline{\boxed{\pink{\bf F_2 = k \dfrac{q_2 q_1}{d^2}}}}$

$\pink{\sf Now, \: for \: \dfrac{F_1}{F_2}:}$

• $\sf F_1 = k \dfrac{q_1 q_2}{d^2}$
• $\sf F_2 = k \dfrac{q_2 q_1}{d^2}$

$\sf : \implies \dfrac{F_1}{F2} = \dfrac{k \dfrac{q_1 q_2}{d^2}}{k \dfrac{q_2 q_1}{d^2}}$

Now,

• q₁ = 1 micro C
• q₂ = 5 micro C

$\sf : \implies \dfrac{F_1}{F2} = \dfrac{k \dfrac{1 \times 5}{d^2}}{k \dfrac{5 \times 1}{d^2}}$

$\sf : \implies \dfrac{F_1}{F2} = \cancel{\dfrac{k \dfrac{5}{d^2}}{k \dfrac{5}{d^2}}}$

$\sf : \implies \dfrac{F_1}{F2} = \dfrac{1}{1}$

$\sf : \implies \dfrac{F_1}{F2} = 1$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \large \underline{\boxed{\pink{\bf \dfrac{F_1}{F2} = 1}}}$