Solveeit Logo
Login

Question

Question: How many electrons should be removed from a coin of mass \[1.6\;{\rm{g}}\], so that it may float in ...

How many electrons should be removed from a coin of mass 1.6  g1.6\;{\rm{g}}, so that it may float in an electric field of intensity 109  N/C{10^9}\;{\rm{N/C}} directed upward?
(1). 9.8×1079.8 \times {10^7}
(2). 9.8×1059.8 \times {10^5}
(3). 9.8×1039.8 \times {10^3}
(4). 9.8×1019.8 \times {10^1}

Explanation

Solution

The above problem can be resolved by applying the formula for the linear force acting on the coin. This linear force is nothing but the gravitational force and the electrostatic force acting on the coin's electrons. Then the equations of force are balanced, assuming the equilibrium condition, where the variable n denotes the desired number of electrons to remove for the coin.

Complete step by step answer:
Given:
The mass of the coin is, m=1.6  g=1.6  g×1  kg1000  g=1.6×10−3  kgm = 1.6\;{\rm{g}} = 1.6\;{\rm{g}} \times \dfrac{{1\;{\rm{kg}}}}{{1000\;{\rm{g}}}} = 1.6 \times {10^{ - 3}}\;{\rm{kg}}
The magnitude of electric field intensity is, E=109  N/CE = {10^9}\;{\rm{N/C}}.
As during floating the electrostatic force on the electrons of the coin is balanced by the overall linear force being influenced on the coin. Then, the relation for the linear force is given as,
F=mg...............(1)F = mg...............\left( 1 \right)
Here, g is the gravitational acceleration and its value is 9.8  m/s29.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}.
And the electrostatic force is given as,
Fe=neE....................(2){F_e} = neE....................\left( 2 \right)
Here, n denotes the number of electrons removed and e is the magnitude of charge on electrons and its value is 1.6×10−19  C1.6 \times {10^{ - 19}}\;{\rm{C}}.
Solve by equating the equation 1 and 2 as,

F = {F_e}\\\ mg = neE\\\ n = \dfrac{{mg}}{{eE}} \end{array}$$ Further solving as, $$\begin{array}{l} n = \dfrac{{6 \times {{10}^{ - 3}}\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}{{1.6 \times {{10}^{ - 19}}\;{\rm{C}} \times {{10}^9}\;{\rm{N/C}}}}\\\ n = 9.8 \times {10^7} \end{array}$$ Therefore, the amount of electrons removed is $$9.8 \times {10^7}$$ and option A is correct. **Note:** To resolve the given problem, try to understand that the concept of equilibrium of forces and all the necessary forces that could affect the motion of the coin. This involves two basic forces, namely the gravitational force and the electrostatic force. These forces are acting in a specific direction to provide the desired motion of the coin.