A proton and an electron are placed 1.6 cm apart in free space. Find the magnitude of electrostatic force between them. The nature of this force.

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$9 \times 10^{- 25} r e p u l s i o n$

$90 \times 10^{- 25} r e p u l s i o n$

$9 \times 10^{- 25} N; a t t r a c t i v e$

$9 \times 10^{25} a t t r a c t i v e$

answer is C.

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Detailed Solution

Given data: a proton and an electron are placed 1.6cm apart in free space.

Concept used: coulomb’s law

Detailed solution:

Charge of proton= $1.6 \times 10^{- 19} C$

Charge of electron= $1.6 \times 10^{- 19} C$

Distance r= $1.6 \times 10^{- 2} m$

Force ,

$F = \frac{k q_{1} q_{2}}{r^{2}}$

$F = 9 \times 10^{- 25} N$

Hence, option(C) is the solution.

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