91Ó°ÊÓ

Quantisation of charge states that the net charge on a charged object is quantised in terms of electronic charge.

According to quantisation of charge, the net charge is given as:

\(Q = ne\) … (i)

Here, n is the number of electrons and e is the electronic charge.

The magnitude of the attractive force is, \(F = 17\;{\rm{mN}} = 17 \times {10^{ - 3}}\;{\rm{N}}\).

The distance between the balls is, \(d = 24\;{\rm{cm}} = 0.24\;{\rm{m}}\).

Since Q charge is transferred from the first ball to the second ball, the charge on the first ball will be Q and on the second ball, the charge will be \( - Q\).

The expression for the magnitude of the force between two charged balls is,

\(\begin{aligned}{c}F = k\frac{{\left| {Q \times \left( { - Q} \right)} \right|}}{{{d^2}}}\\{Q^2} = \frac{{F \times {d^2}}}{k}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}{Q^2} = \frac{{\left( {17 \times {{10}^{ - 3}}\;{\rm{N}}} \right) \times {{\left( {0.24\;{\rm{m}}} \right)}^2}}}{{9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}} \cdot {{\rm{C}}^{{\rm{ - 2}}}}}}\\Q = \sqrt {10.88 \times {{10}^{ - 14}}} \;{\rm{C}}\\Q = 3.3 \times {10^{ - 7}}\;{\rm{C}}\end{aligned}\)

From equation (i), the number of electrons transferred is given as:

\(n = \frac{Q}{e}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}n = \frac{{3.30 \times {{10}^{ - 7}}\;{\rm{C}}}}{{1.6 \times {{10}^{ - 19}}\;{\rm{C}}}}\\n \approx 2.1 \times {10^{12}}\end{aligned}\)

Thus, the number of transferred electrons is \(2.1 \times {10^{12}}\).