Electrostatics

JEE Main level, single correct, medium

29 of 311

Question 29 of 311

Sixty four conducting drops each of radius 0.02m0.02m and each carrying a charge of 5μC5\mu C are combined to form a bigger drop. The ratio of surface density of bigger drop to the smaller drop will be

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Answer: (B) 4:14:1

Total volume will be constant. Therefore, n(4π3r3)=4π3R3⇒6413r=R⇒R=4rn\left(\dfrac{4\pi }{3}{r}^{3}\right)=\dfrac{4\pi }{3}{R}^{3}\Rightarrow {64}^{\dfrac{1}{3}}r=R\Rightarrow R=4r
Final surface charge density σ′=nσ04πr24πR2=64×σ0r216r2⇒σ′σ0=41{\sigma }^{'}=\dfrac{n{\sigma }_{0}4\pi {r}^{2}}{4\pi {R}^{2}}=\dfrac{64\times {\sigma }_{0}{r}^{2}}{16{r}^{2}}\Rightarrow \dfrac{{\sigma }^{'}}{{\sigma }_{0}}=\dfrac{4}{1}

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