Two charges each of $100$ micro coulomb are separated in a medium of relative permittivity $2$ by a distance of $5 c m$ . The force between them is :

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Answer 1

We know force between two charges in a medium separated by a distance 'r'
$F = \frac{( q _{1} ) ( q _{2} )}{4 π ε r ^{2}}$

where, $ϵ_{r} = \frac{ϵ}{ϵ _{o}}$
given, $ϵ_{r} = 2$ $⟹ ϵ = 2 \times ϵ_{o}$

$F = \frac{( 1 0 0 \times 1 0 ^{- 6} ) ( 1 0 0 \times 1 0 ^{- 6} ) \times 9 \times 1 0 ^{9}}{2 \times ( 5 \times 1 0 ^{- 2} ) ^{2}}$

$F = \frac{9 0}{2 \times 2 5 \times 1 0 ^{- 4}}$

$F = \frac{9 0 \times 1 0 0 \times 1 0 ^{2}}{5 0}$

$F = 1 \cdot 8 \times 1 0^{4} N$

Answer 2

We know force between two charges in a medium separated by a distance 'r'

F = \frac{( q _{1} ) ( q _{2} )}{4 π ε r ^{2}}

where,

ϵ_{r} = \frac{ϵ}{ϵ _{o}}

given,

ϵ_{r} = 2

⟹ ϵ = 2 \times ϵ_{o}

F = \frac{( 1 0 0 \times 1 0 ^{- 6} ) ( 1 0 0 \times 1 0 ^{- 6} ) \times 9 \times 1 0 ^{9}}{2 \times ( 5 \times 1 0 ^{- 2} ) ^{2}}

F = \frac{9 0}{2 \times 2 5 \times 1 0 ^{- 4}}

F = \frac{9 0 \times 1 0 0 \times 1 0 ^{2}}{5 0}

F = 1 \cdot 8 \times 1 0^{4} N