Two metallic spheres of radii 1 cm and 3 cm are given charges of − 1 × 1 0 − 2 C and 5 × 1 0 − 2 C , respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is

1 cm k x = 3 cm k ( Q − x ) 3 x = Q − x ⇒ 4 x = Q x = 4 Q = 4 4 × 1 0 − 2 C = 1 × 1 0 − 2 Q ′ = Q − x = 3 × 1 0 − 2 C

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PhysicsElectrostaticsPotential & potential energyMedium2 minPYQ_2010

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Two metallic spheres of radii $1 cm$ and $3 cm$ are given charges of $- 1 \times 1 0^{- 2} C$ and $5 \times 1 0^{- 2} C$ , respectively. If these are connected by a conducting wire, the final charge on the bigger sphere is

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Answer:

Solution:

$\frac{k x}{1 cm} = \frac{k ( Q - x )}{3 cm}$ $3 x = Q - x \Rightarrow 4 x = Q$ $x = \frac{Q}{4} = \frac{4 \times 1 0 ^{- 2}}{4} C = 1 \times 1 0^{- 2}$ $Q^{'} = Q - x = 3 \times 1 0^{- 2} C$

Stream:BITSATSubject:PhysicsTopic:ElectrostaticsSubtopic:Potential & potential energy

⏱ 2mℹ️ Source: PYQ_2010

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