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16-13P

Expert-verifiedFound in: Page 443

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**(II) A charge Q is transferred from an initially uncharged plastic ball to an identical ball 24 cm away. The force of attraction is then 17 mN. How many electrons were transferred from one ball to the other?**

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

**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 \(2.1 \times {10^{12}}\).

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