Two isolated metallic solid spheres of radii $R$ and $2 R$ are charged such that both have same charge density $\sigma$. The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is $\sigma^{\prime}$ The ratio $\frac{\sigma^{\prime}}{\sigma}$ is :

Updated On: Sep 24, 2024

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The Correct Option is C

Two isolated metallic solid spheres of radii R and 2R are charged

\frac{Q _{1}^{'}}{4 π ε _{0} R} = \frac{Q _{2}^{'}}{4 π ε _{0} ( 2 R )}

∴ Q_{2}^{'} = 2 Q_{1}^{'}

Q_{1}^{'} + Q_{2}^{'} = Q_{1} + Q_{2}

∴ \frac{Q _{2}^{'}}{2} + Q_{2}^{'} = 20 π R^{2} σ

\frac{3}{2} Q_{2}^{'} = 20 π R^{2} σ

∴ \frac{Q _{2}^{'}}{4 π ( 2 R ) ^{2}} = \frac{2}{3} \cdot \frac{20 π R ^{2} σ}{16 π R ^{2}}

∴ \frac{σ ^{'}}{σ} = \frac{5}{6}

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Electric Field is the electric force experienced by a unit charge.

The electric force is calculated using the coulomb's law, whose formula is:

$F=k\dfrac{|q_{1}q_{2}|}{r^{2}}$

While substituting q₂ as 1, electric field becomes:

$E=k\dfrac{|q_{1}|}{r^{2}}$

SI unit of Electric Field is V/m (Volt per meter).

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