The force between two small charged spheres having charges of 1 × 1 0 − 7 C and 2 × 1 0 − 7 C placed 20 cm apart in air is

Here, q 1 = 1 × 1 0 − 7 C , q 2 and 2 × 1 0 − 7 C , r = 20 cm = 20 × 1 0 − 2 m F = 4 π ε 0 r 2 q 1 q 2 = ( 20 × 1 0 − 2 ) 2 9 × 1 0 9 × 1 × 1 0 − 7 × 2 × 1 0 − 7 = 4.5 × 1 0 − 3 N

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PhysicsElectrostaticsElectric Field & ForceEasy2 minPYQ_2023

PhysicsEasysingle choice

The force between two small charged spheres having charges of $1 \times 1 0^{- 7} C$ and $2 \times 1 0^{- 7} C$ placed $20 cm$ apart in air is

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Solution:

Here, $q_{1} = 1 \times 1 0^{- 7} C, q_{2}$ and $2 \times 1 0^{- 7} C$ , $r = 20 cm = 20 \times 1 0^{- 2} m$
$F = \frac{q _{1} q _{2}}{4 π ε _{0} r ^{2}} = \frac{9 \times 1 0 ^{9} \times 1 \times 1 0 ^{- 7} \times 2 \times 1 0 ^{- 7}}{( 20 \times 1 0 ^{- 2} ) ^{2}}$
$= 4.5 \times 1 0^{- 3} N$

Stream:BITSATSubject:PhysicsTopic:ElectrostaticsSubtopic:Electric Field & Force

⏱ 2mℹ️ Source: PYQ_2023

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