Two charged spheres of radius $R_{1}$ and $R_{2}$ respectively, are charged and joined by a wire. The ratio of electric field of the spheres is

\frac{R_{1}}{R_{2}}

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\frac{R_{2}}{R_{1}}

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\frac{R_{1}^{2}}{R_{2}^{2}}

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\frac{R_{2}^{2}}{R_{1}^{2}}

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Solution

The correct option is B $\frac{R_{2}}{R_{1}}$
Two charged conducting sphere of radii r₁ and r₂ respectively connected to each other by a wire.
so, capacitance of 1st sphere , $C ₁ = 4 π ε ₀ r ₁$
capacitance of 2nd sphere , $C ₂ = 4 π ε ₀ r ₂$
After connecting , transfer of charge from higher potential sphere to lower potential sphere and finally, potential of both sphere will be equal .Let final potential is $V$
Now, charge at 1st sphere, $Q ₁ = C ₁ V = 4 π ε ₀ r ₁ V$
charge at 2nd sphere , $Q ₂ = C ₂ V = 4 π ε ₀ r ₂ V$
so, ratio of charge $= Q ₁ / Q ₂ = 4 π ε ₀ r ₁ V / 4 π ε ₀ r ₂ V = r ₁ / r ₂ - - - - - - - (1)$
Now, ratio of electric field $= E ₁ / E ₂ = K Q ₁ / r ₁ ² / K Q ₂ / r ₂ ² = Q ₁ r ₂ ² / Q ₂ r ₁ ²$
$E ₁ / E ₂ = Q ₁ / Q ₂ r ₂ ² / r ₁ ²$
Put equation (1)
$E ₁ / E ₂ = r ₂ / r ₁$
Hence, the option $B$ is the correct answer.

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