Question:
A charged oil drop is suspended in a uniform electric field of $3 \times 10^{4}\, V / m$ so that it neither falls nor rises. The charge on the drop will be (take the mass of the drop is $9.9 \times 10^{-15} \,kg$ and $g=10 \,ms ^{-2}$ )
A charged oil drop is suspended in a uniform electric field of $3 \times 10^{4}\, V / m$ so that it neither falls nor rises. The charge on the drop will be (take the mass of the drop is $9.9 \times 10^{-15} \,kg$ and $g=10 \,ms ^{-2}$ )
Updated On: Jan 18, 2023
- $ 3.3\times {{10}^{18}}C $
- $ 3.2\times {{10}^{-18}}C $
- $ 1.6\times {{10}^{-18}}C $
- $ 4.8\times {{10}^{-18}}C $
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The Correct Option is A
Solution and Explanation
Because the drop neither falls nor rises, so it remains stationary, then
For equilibrium position, the electric farce $q E$ will be equal to the weight of the drop.
So, $q E=m g q=\frac{m g}{E}$
$=\frac{\left(9.9 \times 10^{-15}\right) \times 10}{3 \times 10^{4}} $
$=3.3 \times 10^{-18} C$
For equilibrium position, the electric farce $q E$ will be equal to the weight of the drop.
So, $q E=m g q=\frac{m g}{E}$
$=\frac{\left(9.9 \times 10^{-15}\right) \times 10}{3 \times 10^{4}} $
$=3.3 \times 10^{-18} C$
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Concepts Used:
Coulomb’s Law
In 1785, french physicist Charles Augustin de Coulomb coined a tangible relationship in mathematical form between two bodies that have been electrically charged. He represented an equation for the force causing the bodies to attract or repel each other which is commonly known as Coulomb’s law or Coulomb’s inverse-square law.
As per Coulomb’s law, the force of attraction or repulsion between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. It acts along the line joining the two charges regarded to be point charges.
Coulomb’s Law has an abundant application to modern life, from Xerox machines to laser printers, to powder coating.