Question

A plastic ball \( P \) of mass \( 3.2 \times 10^{-15} \, \mathrm{kg} \) is suspended between two horizontal parallel charged plates in a balanced state. How many electrons on the ball will be increased or decreased? \((g = 10 \, \mathrm{m/s}^2)\)

Hint:

To calculate the number of electrons added or removed, use \( n = \frac{q}{e} \), where \( q \) is the charge and \( e \) is the charge of an electron.

Answer 1

The plastic ball \( P \) is in a balanced state, meaning the gravitational force \( mg \) acting downwards is balanced by the electric force \( qE \) acting upwards. \[ mg = qE \quad \Rightarrow \quad q = \frac{mg}{E}. \] The electric field \( E \) between the plates is given by: \[ E = \frac{\text{Potential Difference}}{\text{Distance}} = \frac{1000}{0.02} = 50000 \, \mathrm{V/m}. \] Substituting values: \[ q = \frac{(3.2 \times 10^{-15}) (10)}{50000} = 6.4 \times 10^{-19} \, \mathrm{C}. \] The charge of a single electron is \( e = 1.6 \times 10^{-19} \, \mathrm{C} \). The number of electrons \( n \) added to the ball can be calculated as: \[ n = \frac{q}{e} = \frac{6.4 \times 10^{-19}}{1.6 \times 10^{-19}} = 4. \] Thus, the ball must gain \( 4 \, \mathrm{electrons} \).