JEE Advanced · Physics · 17. Electrostatics
A tiny spherical oil drop carrying a net charge \(q\) is balanced in still air with a vertical uniform electric field of strength \(\frac{81 \pi}{7} \times 10^5 \mathrm{Vm}^{-1}\). When the field is switched off, the drop is observed to fall with terminal velocity \(2 \times 10^{-3} \mathrm{~ms}^{-1}\). Given \(g=9.8 \mathrm{~ms}^{-2}\), viscosity of the air \(=1.8 \times 10^{-5} \mathrm{Ns} \mathrm{m}^{-2}\) and the density of oil \(=900 \mathrm{~kg} \mathrm{~m}^{-3}\), the magnitude of \(q\) is
- A \(1.6 \times 10^{-19} \mathrm{C}\)
- B \(3.2 \times 10^{-19} \mathrm{C}\)
- C \(4.8 \times 10^{-19} \mathrm{C}\)
- D \(8.0 \times 10^{-19} \mathrm{C}\)
Answer & Solution
Correct Answer
(D) \(8.0 \times 10^{-19} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& q E=m g \\
& \quad 6 \pi \eta r v=m g \\
& \frac{4}{3} \pi r^3 \rho g=m g \\
& \therefore \quad r=\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3}
\end{aligned}
\)
Substituting the value of \(r\) in Eq. (ii) we get,
\(
6 \pi \eta v\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3}=m g
\)
or \(\quad(6 \pi \eta v)^3\left(\frac{3 m g}{4 \pi \rho g}\right)=(m g)^3\)
Again substituting \(m g=q E\) we get,
\(
(q E)^2=\left(\frac{3}{4 \pi \rho g}\right)(6 \pi \eta v)^3
\)
or \(\quad q E=\left(\frac{3}{4 \pi \rho g}\right)^{1 / 2}(6 \pi \eta v)^{3 / 2}\)
\(
\therefore \quad q=\frac{1}{E}\left(\frac{3}{4 \pi \rho g}\right)^{1 / 2}(6 \pi \eta v)^{3 / 2}
\)
Substituting the values we get ,
\(
\begin{aligned}
q= & \frac{7}{81 \pi \times 10^5} \sqrt{\frac{3}{4 \pi \times 900 \times 9.8} \times 216 \pi^3} \\
& \times \sqrt{\left(1.8 \times 10^{-5} \times 2 \times 10^{-3}\right)^3} \\
& =8.0 \times 10^{-19} \mathrm{C}
\end{aligned}
\)
\(\therefore\) correct option is \((\mathrm{d})\).
\begin{aligned}
& q E=m g \\
& \quad 6 \pi \eta r v=m g \\
& \frac{4}{3} \pi r^3 \rho g=m g \\
& \therefore \quad r=\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3}
\end{aligned}
\)
Substituting the value of \(r\) in Eq. (ii) we get,
\(
6 \pi \eta v\left(\frac{3 m g}{4 \pi \rho g}\right)^{1 / 3}=m g
\)
or \(\quad(6 \pi \eta v)^3\left(\frac{3 m g}{4 \pi \rho g}\right)=(m g)^3\)
Again substituting \(m g=q E\) we get,
\(
(q E)^2=\left(\frac{3}{4 \pi \rho g}\right)(6 \pi \eta v)^3
\)
or \(\quad q E=\left(\frac{3}{4 \pi \rho g}\right)^{1 / 2}(6 \pi \eta v)^{3 / 2}\)
\(
\therefore \quad q=\frac{1}{E}\left(\frac{3}{4 \pi \rho g}\right)^{1 / 2}(6 \pi \eta v)^{3 / 2}
\)
Substituting the values we get ,
\(
\begin{aligned}
q= & \frac{7}{81 \pi \times 10^5} \sqrt{\frac{3}{4 \pi \times 900 \times 9.8} \times 216 \pi^3} \\
& \times \sqrt{\left(1.8 \times 10^{-5} \times 2 \times 10^{-3}\right)^3} \\
& =8.0 \times 10^{-19} \mathrm{C}
\end{aligned}
\)
\(\therefore\) correct option is \((\mathrm{d})\).
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