KCET · Physics · Electrostatics
A small oil drop of mass \(10^{-6} \mathrm{~kg}\) is hanging in at rest between two plates separated by \(1 \mathrm{~mm}\) having a potential difference of \(500 \mathrm{~V}\). The charge on the drop is \(\left(g=10 \mathrm{~ms}^{-2}\right.\) )
- A \(2 \times 10^{-9} \mathrm{C}\)
- B \(2 \times 10^{-11} \mathrm{C}\)
- C \(2 \times 10^{-6} \mathrm{C}\)
- D \(2 \times 10^{-9} \mathrm{C}\)
Answer & Solution
Correct Answer
(B) \(2 \times 10^{-11} \mathrm{C}\)
Step-by-step Solution
Detailed explanation
Given, the drop rests between the two plates
\(\therefore q E=m g \)
\( \text {or } \quad q \frac{V}{r}=m g \quad\left(\because E=\frac{V}{r}\right) \)
\( \Rightarrow q=\frac{m g r}{V} \)
\( \text {here, } m=10^{-6} \mathrm{~kg}, g=10 \mathrm{~m} / \mathrm{s}^{2}, r=1 \mathrm{~mm}=10^{-3} \mathrm{~m} \)
and
Substituting all the values, we get
\( q=\frac{10^{-6} \times 10 \times 10^{-3}}{500}, q=2 \times 10^{-11} \mathrm{C}\)
\(\therefore q E=m g \)
\( \text {or } \quad q \frac{V}{r}=m g \quad\left(\because E=\frac{V}{r}\right) \)
\( \Rightarrow q=\frac{m g r}{V} \)
\( \text {here, } m=10^{-6} \mathrm{~kg}, g=10 \mathrm{~m} / \mathrm{s}^{2}, r=1 \mathrm{~mm}=10^{-3} \mathrm{~m} \)
and
Substituting all the values, we get
\( q=\frac{10^{-6} \times 10 \times 10^{-3}}{500}, q=2 \times 10^{-11} \mathrm{C}\)
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