Question

For an experiment, a colleague of yours says he smeared toner particles uniformly over the surface of a sphere 1.0 m in diameter and then measured an electric field of \({\bf{5000 N/C}}\) near its surface. (a) How many toner particles (Example 16–6) would have to be on the surface to produce these results? (b) What is the total mass of the toner particles?

Short answer

1. The number of toner particles is\(4.34 \times {10^{10}}\).
2. The total mass of the toner particles is \(3.91 \times {10^{ - 5}}\,{\rm{kg}}\).

Step-by-step solution

Understanding the electric field of the toner particles

The electric field of the toner particles is dependent on the charge of the particles and the separation distance between them.The total charge of the toner particles is the product of the number of toner particles and the charge on each toner particle.

Identification of given data

The given data can be listed below as:

• The charge of the toner particles is\(q = 20e\).
• The total charge of the toner particles is\(Q = Nq\).
• The number of toner particles is N.
• The mass of the toner particles is\(m = 9 \times {10^{ - 16}}{\rm{ kg}}\).
• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).
• The value of the electric field is\(E = 5000{\rm{ N/C}}\).
• The diameter of the sphere is\(d = 1{\rm{ m}}\).
• The charge on an electron is\(e = 1.6 \times {10^{ - 19}}{\rm{ C}}\).
• The Coulomb’s constant value is \(k = 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}\).

(a) Determination of the total number of toner particles

Identical charged particles are smeared uniformly over the surface of the sphere. Every toner particle has the same charge of 20e.

The electric field can be expressed as:

\(\begin{aligned}{c}E &= k\frac{Q}{{{{\left( r \right)}^2}}}\\E &= k\frac{{Nq}}{{{{\left( {\frac{d}{2}} \right)}^2}}}\\E &= \frac{{4kNq}}{{{d^2}}}\\N &= \frac{{E{d^2}}}{{4kq}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}N &= \frac{{5000{\rm{ N/C}} \times {{\left( {1{\rm{ m}}} \right)}^2}}}{{4\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}} \right) \times 20e}}\\ &= \frac{{5000{\rm{ N/C}} \times {{\left( {1{\rm{ m}}} \right)}^2}}}{{4\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}} \right) \times 20 \times 1.6 \times {{10}^{ - 19}}{\rm{ C}}}}\\ &= \frac{{5000}}{{1.152 \times {{10}^{ - 7}}}}\\ &= 4.34 \times {10^{10}}\end{aligned}\)

Thus, the number of toner particles is \(4.34 \times {10^{10}}\).

(b) Determination of the total mass of the toner particles

The total mass of the toner particles can be expressed as:

\(M = Nm\)

Substitute the values in the above equation.

\(\begin{aligned}{c}M &= 4.34 \times {10^{10}} \times 9 \times {10^{ - 16}}{\rm{ kg}}\\ &= 3.91 \times {10^{ - 5}}\,{\rm{kg}}\end{aligned}\)

Thus, the total mass of the toner particles is \(3.91 \times {10^{ - 5}}\,{\rm{kg}}\).