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Standard XII

Physics

Coulomb's Law - Mathematical Form

Two pith ball...

Question

Two pith balls each of mass $1 g$ and carrying a charge 1 $μ$ C are attached to the ends of silk threads $1 m$ long, the other ends of which are attached to some fixed point, in a gravity free space. The force between them is :

9.8 \times 10^{- 2} N

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9.0 \times 10^{- 3} N

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4.5 \times 10^{- 2} N

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0.9 \times 10^{- 3} N

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Solution

The correct option is B $9.0 \times 10^{- 3} N$

As they are in a gravity free space so the force between them is only the electrostatic force. i.e,

$F = \frac{q^{2}}{4 π ϵ_{0} r^{2}} = \frac{9 \times 10^{9} (10^{- 6})^{2}}{1^{2}} = 9 \times 10^{- 3} N$

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Two pith balls each of mass 1g and carrying a charge 1μC are attached to the ends of silk threads 1m long, the other ends of which are attached to some fixed point, in a gravity free space. The force between them is :

The correct option is B 9.0×10−3N As they are in a gravity free space so the force between them is only the electrostatic force. i.e, F=q24πϵ0r2=9×109(10−6)212=9×10−3N

Two pith balls each of mass $1 g$ and carrying a charge 1 $μ$ C are attached to the ends of silk threads $1 m$ long, the other ends of which are attached to some fixed point, in a gravity free space. The force between them is :

The correct option is B $9.0 \times 10^{- 3} N$

As they are in a gravity free space so the force between them is only the electrostatic force. i.e,

$F = \frac{q^{2}}{4 π ϵ_{0} r^{2}} = \frac{9 \times 10^{9} (10^{- 6})^{2}}{1^{2}} = 9 \times 10^{- 3} N$