Question

Packing material made of pieces of foamed polystyrene can easily become charged and stick to each other. Given that the density of this material is about \({\bf{35 kg/}}{{\bf{m}}^{\bf{3}}}\), estimate how much charge might be on a 2.0-cm-diameter foamed polystyrene sphere, assuming the electric force between two spheres stuck together is equal to the weight of one sphere.

Short answer

The magnitude of the charge is \(8 \times {10^{ - 9}}{\rm{ C}}\).

Step-by-step solution

Understanding the forces acting on the spheres

The electric force between the two spheres is due to the charge on them. The weight of the sphere acts in the downward direction. The electric force depends on the sphere’s charges and the separation between the centers of the spheres. The distance between the centers is equal to the diameter of the spheres.

Identification of given data

The given data can be listed below as:

• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).
• The diameter of the sphere is\(d = 2{\rm{ cm}}\left( {\frac{{1{\rm{ m}}}}{{100{\rm{ cm}}}}} \right) = 0.02{\rm{ m}}\).
• The density of the material is\(\rho = {\rm{35 kg/}}{{\rm{m}}^3}\).
• The Coulomb’s law constant value is \(k = 9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}\).

Representation of the spheres

The diagram of the two spheres can be shown as:

Here, r is the radius of spheres (1) and (2) and d is the distance between the two charges of equal magnitude (Q). This separation is equal to the diameter of one sphere.

Determination of the magnitude of the charge

According to Coulomb’s law, the force acting between the two charged spheres can be expressed as:

\(F = k\frac{{\left| {QQ} \right|}}{{{{\left( {2r} \right)}^2}}}\)

\(F = k\frac{{{Q^2}}}{{{{\left( d \right)}^2}}}\) … (i)

Here, both charges are of equal magnitude (Q). The distance between both charges is (2r), which is equal to the diameter (d) of one sphere.

The weight of one sphere can be expressed as:

\(\begin{aligned}{c}W &= mg\\W &= \rho Vg\\W &= \rho \left( {\frac{{4\pi {r^3}}}{3}} \right)g\end{aligned}\)

\(W = \rho \times \frac{4}{3}\pi {\left( {\frac{d}{2}} \right)^3} \times g\) … (ii)

Here, m is the mass of one sphere and V is the volume of the sphere.

Equate equations (i) and (ii), and evaluate the magnitude of the charge.

\(\begin{aligned}{c}F &= W\\k\frac{{{Q^2}}}{{{{\left( d \right)}^2}}} &= \rho \times \frac{4}{3}\pi {\left( {\frac{d}{2}} \right)^3} \times g\\{Q^2} &= \frac{4}{3} \times \frac{1}{8} \times \pi \times \frac{{\rho g{d^5}}}{k}\\Q &= \sqrt {\frac{1}{6}\left( {\frac{{\pi \rho g{d^5}}}{k}} \right)} \end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{c}Q &= \sqrt {\frac{1}{6}\left( {\frac{{\pi \times {\rm{35 kg/}}{{\rm{m}}^3} \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times {{\left( {0.02{\rm{ m}}} \right)}^5}}}{{9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}}}} \right)} \\ &= \sqrt {\frac{1}{6}\left( {\frac{{3.452 \times {{10}^{ - 6}}{\rm{ kg}} \cdot {{\rm{m}}^3}{\rm{/}}{{\rm{s}}^2}\left( {\frac{{1{\rm{ N}}}}{{1{\rm{ kg}} \cdot {\rm{m/}}{{\rm{s}}^2}}}} \right)}}{{9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}}}} \right)} \\ &= \sqrt {\frac{1}{6} \times \left( {3.835 \times {{10}^{ - 16}}} \right){\rm{ }}{{\rm{C}}^2}} \\ \approx 8 \times {10^{ - 9}}{\rm{ C}}\end{aligned}\)

Thus, the magnitude of the charge is\(8 \times {10^{ - 9}}{\rm{ C}}\).