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Standard XII

Physics

Understanding the Electrostatic Model

Point charges...

Question

Point charges $Q_{1} = 5 m - C$ (milli-coulomb) and $Q_{2} = 6 m - C$ (milli-coulomb) are at a separation of 100m from each other. What is the magnitude of force on any one of the charge?

27N

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25N on Q₁ and 24N on Q₂

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23N

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not enough information is given in the question.

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Solution

The correct option is A
27N

$F = \frac{K q_{1} q_{2}}{r^{2}} = \frac{9 \times 10^{9} \times 5 \times 10^{- 3} \times 6 \times 10^{- 3}}{10^{4}}$

$= 27 N$

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Similar questions

Point charges $Q_{1} = 5 m C$ (millicoulomb) and $Q_{2} = 6 m C$ (millicoulomb) are at a separation of 100m from each other. What is the magnitude of force on any one of the charge?

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Charges Q₁ and Q₂ are near each other. We have no information of their position, nature and magnitude, all we know is Q₂ experiences a force of 5N to the right. What can we infer for sure from this?

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Point charges Q1=5m−C (milli-coulomb) and Q2=6m−C (milli-coulomb) are at a separation of 100m from each other. What is the magnitude of force on any one of the charge?

The correct option is A 27N F=Kq1q2r2=9×109×5×10−3×6×10−3104 =27N

Point charges $Q_{1} = 5 m - C$ (milli-coulomb) and $Q_{2} = 6 m - C$ (milli-coulomb) are at a separation of 100m from each other. What is the magnitude of force on any one of the charge?

The correct option is A
27N

$F = \frac{K q_{1} q_{2}}{r^{2}} = \frac{9 \times 10^{9} \times 5 \times 10^{- 3} \times 6 \times 10^{- 3}}{10^{4}}$

$= 27 N$