Electrostatics

JEE Main level, single correct, medium

144 of 311

Question 144 of 311

A charged oil drop is suspended in a uniform field of 3×104 V/m3 \times 10^4 \mathrm{~V} / \mathrm{m} so that it neither falls nor rises. The charge on the drop will be (take the mass of the charge =9.9×10−15 kg=9.9 \times 10^{-15} \mathrm{~kg} and g=\mathrm{g}= 10 m/s210 \mathrm{~m} / \mathrm{s}^2 )

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Answer: (A) 3.3×10−18C3.3 \times 10^{-18} \mathrm{C}

Since ball is hanging in equilibrium, force by gravity is balanced by electric force.
qE=mg⇒q=m×gE⇒9.9×10−15×103×104∴q=3.3×10−18C\begin{aligned} & \mathrm{qE}=\mathrm{mg} \\ & \Rightarrow \mathrm{q}=\frac{\mathrm{m} \times \mathrm{g}}{\mathrm{E}} \\ & \Rightarrow \frac{9.9 \times 10^{-15} \times 10}{3 \times 10^4} \\ & \therefore \mathrm{q}=3.3 \times 10^{-18} \mathrm{C} \end{aligned}

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