The charge of one electron is \( e = 1.6 \times 10^{-19} \, \text{C} \).
When electrons are removed from a neutral metal plate, the charge on the plate becomes positive because the removal of negatively charged particles leads to an excess of positive charge.
The total charge removed from the metal plate can be calculated by: \[ Q = n \times e \] where:
- \( Q \) is the total charge,
- \( n = 10^{19} \) is the number of electrons removed,
- \( e = 1.6 \times 10^{-19} \, \text{C} \) is the charge of one electron. Substituting the values: \[ Q = 10^{19} \times 1.6 \times 10^{-19} \, \text{C} = 1.6 \, \text{C} \]
Thus, the charge on the plate is \( +1.6 \, \text{C} \). Therefore, the correct answer is: \[ \text{(2) } +1.6 \, \text{C} \]
Four capacitors each of capacitance $16\,\mu F$ are connected as shown in the figure. The capacitance between points A and B is __ (in $\mu F$)
Two point charges M and N having charges +q and -q respectively are placed at a distance apart. Force acting between them is F. If 30% of charge of N is transferred to M, then the force between the charges becomes:
If the ratio of lengths, radii and Young's Moduli of steel and brass wires in the figure are $ a $, $ b $, and $ c $ respectively, then the corresponding ratio of increase in their lengths would be: