Two point charges q1 and q2 are 3 m apart and their combined charge is 20- C- If one repels the other with a force of 0-075 N- what are two charges-

Let q_1q_2 be two chargesGiven q_1+q_2=20× 10^ 6C From Coulomb's law, F=14πϵ_0q_1q_2r^2 ⇒ 0.75=9× 10^9×q_1q_2(3)^2 ⇒ q_1q_2=0.075× 99× ...

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Standard IX

Physics

Second Law of Motion

Two point cha...

Question

Two point charges $q_{1}$ and $q_{2}$ are 3 m apart and their combined charge is $20 μ C$ . If one repels the other with a force of 0.075 N, what are two charges?

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Solution

Let $q_{1} q_{2}$ be two charges
Given $q_{1} + q_{2} = 20 \times 10^{- 6} C$
From Coulomb's law, $F = \frac{1}{4 π ϵ_{0}} \frac{q_{1} q_{2}}{r^{2}}$
$\Rightarrow 0.75 = 9 \times 10^{9} \times \frac{q_{1} q_{2}}{(3)^{2}}$
$\Rightarrow q_{1} q_{2} = \frac{0.075 \times 9}{9 \times 10^{9}} = 75 \times 10^{- 12} C^{2}$
$(q_{1} = q_{2})^{2} = (q_{1} + q_{2})^{2} - 4 q_{1} q_{2}$
$= (20 \times 10^{- 6})^{2} - 4 \times 75 \times 10^{- 12}$
$= 10^{- 10} C^{2}$
or $q_{1} - q_{2} = 10^{- 5} C = 10 \times 10^{- 6} C$
$(q_{1} + q_{2}) + (q_{1} - q_{2})$
$= (20 \times 10^{- 6}) + (10 \times 10^{- 6})$
$= 30 \times 10^{- 6}$
or $2 q_{1} = 30 \times 10^{- 6} C$
$\Rightarrow q_{1} = 15 \times 10^{- 6} C = 15 μ C$
then, $q_{2} = 20 \times 10^{- 6} - 15 \times 10^{- 6}$
$= 5 \times 10^{- 6} = 5 μ C$

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Two point charges q1 and q2 are 3 m apart and their combined charge is 20μC. If one repels the other with a force of 0.075 N, what are two charges?

Two point charges $q_{1}$ and $q_{2}$ are 3 m apart and their combined charge is $20 μ C$ . If one repels the other with a force of 0.075 N, what are two charges?