Question

Two identical metallic sphere are charged
with 10 and -20 units of charge. If both
the spheres are first brought into contact
with each other and then are placed to
their previous positions, then the ratio of
the force in the two situations will be :

Answer 1

$\underline{\underline{\blacksquare\:\:\:\footnotesize{\red{\textsf{ \textbf{{SOLUTION:-}}}}}}}$

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• The original charges of the two identical metallic sphere are :

• Q₁ = 10 μC = 10 × 10⁻⁶ C
• Q₂ = -20 μC = -20 × 10⁻⁶ C

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• After the spheres are brought into contact with each other, the new charges will be appear as :

$\footnotesize{ \longrightarrow \: \sf Q' _{1} = \dfrac{Q_1 + Q_2}{2} =Q' _{2} }$

$\footnotesize{ \longrightarrow \: \sf Q' _{1} = \dfrac{10 \times {10}^{ - 6} + ( - 20) \times {10}^{ - 6} }{2} =Q' _{2} }$

$\footnotesize{ \longrightarrow \: \sf Q' _{1} = \dfrac{(10 - 20 ) ({10}^{ - 6} ) }{2} =Q' _{2} }$

$\footnotesize{ \longrightarrow \: \sf Q' _{1} = \dfrac{ - 10 \times {10}^{ - 6} }{2} =Q' _{2} }$

$\footnotesize{ \longrightarrow \: \sf Q' _{1} = - 5 \times {10}^{ - 6} \: C=Q' _{2} }$

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Electrostatic force is give by :

$\footnotesize{ \longrightarrow \: \sf F_e =k \dfrac{Q_1 Q_2 }{ {r}^{2} } }$

Where,

$\footnotesize{ \longrightarrow \: \sf F_e \propto Q_1 Q_2}$

Therefore, the ratio of the force in the two situations will be :

$\footnotesize{ \longrightarrow \: \sf \dfrac{F_e}{F_e'} = \dfrac{Q_1 Q_2}{Q_1' Q_2'}}$

$\footnotesize{ \longrightarrow \: \sf \dfrac{F_e}{F_e'} = \dfrac{10 \times {10}^{ - 6 } \times - 20 \times {10}^{ - 6} }{ - 5 \times {10}^{ - 6} \times - 5 \times {10}^{ - 6} }}$

$\footnotesize{ \longrightarrow \: \sf \dfrac{F_e}{F_e'} = \dfrac{10 \times - 20 }{ - 5 \times - 5 }}$

$\footnotesize{ \longrightarrow \: \sf \dfrac{F_e}{F_e'} = \dfrac{ - 20 0}{ 25}}$

$\footnotesize{ \longrightarrow \: \gray{\underline{ \boxed{\orange{\bf \dfrac{F_e}{F_e'} = \dfrac{ -8}{1}}}}}}$

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