Number of electrons removed to acquire charge – Rankers Physics

Properties of Charges: Practice Problem & Solution

The number of electrons that should be removed from a metal coin such that coin acquires a positive charge of \(10^{-10}\text{ C}\) is
\(1.6 \times 10^{-19}\)
\(6.25 \times 10^9\)
\(6.25 \times 10^8\)
\(1.6 \times 10^8\)

Solution Explained:

To solve this problem, we apply the core principles of Properties of Charges. Understanding the underlying formula is key to arriving at the correct answer below:

By quantization of charge, \(q = ne\). Solving for \(n\), we get \(n = \frac{q}{e} = \frac{10^{-10}\text{ C}}{1.6 \times 10^{-19}\text{ C}} = 6.25 \times 10^8\).

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Sixty four identical drops of water having equal charge combine to form a bigger drop. The factor by which potential of bigger drop change in comparison to a small drop is
2. Four electric charges of \(8\ mu\text{C}\), \(5\ mu\text{C}\), \(-3\ mu\text{C}\), \(-10\ mu\text{C}\) are placed at the corners of a square of side \(\sqrt{2}\text{ m}\). The potential at the centre of the square is
3. Assertion (A): The tyres of aircrafts are slightly conducting. Reason (R): If a conductor is connected to ground, the extra charge induced on conductor will flow to ground.  
4. A spherical conductor of radius $10 \text{ cm}$ has a charge of $3.2 \times 10^{-7} \text{ C}$ distributed uniformly. What is the magnitude of electric field at a point $15 \text{ cm}$ from the centre of the sphere? (2020) $(\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2)$
5. Electric field at the equator of a dipole is $E$. If strength and distance is now doubled then the electric field will be: (1998)