Two charged particles are placed at a distance of 1.0 cm apart. What is the minimum possible magnitude of electrostatic force acting on each charge

25 x

10^{- 24}

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23 x

10^{- 21}

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2.3 x

10^{- 24}

dyne

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2.3 x

10^{- 24}

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Solution

The correct option is D 2.3 x $10^{- 24}$ N
$\begin{matrix} F = \frac{1}{4 π \in_{o}} \frac{q_{1} q_{2}}{r^{2}} q_{1} min = q_{2} min = 1.6 \times 10^{- 19} C F min = 2.304 \times 10^{- 24} N \end{matrix}$

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Two charged particles are placed at a distance of 1.0 cm apart. What is the minimum possible magnitude of electrostatic force acting on each charge

The correct option is D 2.3 x 10−24NF=14π∈oq1q2r2q1min=q2min=1.6×10−19CFmin=2.304×10−24N

The correct option is D 2.3 x $10^{- 24}$ N
$\begin{matrix} F = \frac{1}{4 π \in_{o}} \frac{q_{1} q_{2}}{r^{2}} q_{1} min = q_{2} min = 1.6 \times 10^{- 19} C F min = 2.304 \times 10^{- 24} N \end{matrix}$