Question

(a) Common transparent tape becomes charged when pulled from a dispenser. If one piece is placed above another, the repulsive force can be great enough to support the top piece’s weight. Assuming equal point charges (only an approximation), calculate the magnitude of the charge if electrostatic force is great enough to support the weight of a\[{\bf{10}}.{\bf{0}}{\rm{ mg}}\]piece of tape held\[{\bf{1}}.0{\bf{0}}{\rm{ cm}}\]above another. (b) Discuss whether the magnitude of this charge is consistent with what is typical of static electricity.

Short answer

(a) The magnitude of the charges will be \[1.04{\rm{ }}nC\].

(b) The magnitude of the charge will be consistent with the static electricity as long as the tap is pulled from the dispenser.

Step-by-step solution

Electrostatic levitation

The process of using an electric field to levitate (float) a charged object by providing an upwards repulsive force that counteracts the pull of gravity is known as electrostatic force of repulsion.

Magnitude of Charge

(a)

The electrostatic force between two similar point charges q , separated by some distance r is,

\(F = \frac{{K{q^2}}}{{{r^2}}}\)

Here, K is the electrostatic force constant \(\left( {K = 9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\), q is the charges on the tapes, and r is the separation between tapes \(\left( {r = 1.00{\rm{ cm}}} \right)\).

The force of gravity or weight of the tape is,

\(F = mg\)

Here, m is the mass of the tape \(\left( {m = 10{\rm{ }}mg} \right)\), and g is the acceleration due to gravity \(\left( {g = 9.8{\rm{ m/}}{{\rm{s}}^{\rm{2}}}} \right)\).

Since, the electrostatic force supports the weight of the tape. Therefore,

\(mg = \frac{{K{q^2}}}{{{r^2}}}\)

The expression for the charge is,

\(q = \sqrt {\frac{{mg{r^2}}}{K}} \)

Substituting all known values,

\[\begin{aligned} {\rm{q}} &= \sqrt {\frac{{\left( {10{\rm{ }}mg} \right) \times \left( {9.8{\rm{ }}m/{s^2}} \right) \times {{\left( {1.00{\rm{ }}cm} \right)}^2}}}{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right)}}} \\ &= \sqrt {\frac{{\left( {10{\rm{ }}mg} \right) \times \left( {\frac{{{{10}^{ - 6}}{\rm{ }}kg}}{{1{\rm{ }}mg}}} \right) \times \left( {9.8{\rm{ }}m/{s^2}} \right) \times {{\left[ {\left( {1.00{\rm{ }}cm} \right) \times \left( {\frac{{{{10}^{ - 2}}{\rm{ }}m}}{{1{\rm{ }}cm}}} \right)} \right]}^2}}}{{\left( {9 \times {{10}^9}{\rm{ }}N \times {m^2}/{C^2}} \right)}}} \\ &= 1.04 \times {10^{ - 9}}{\rm{ }}C \times \left( {\frac{{1{\rm{ }}nC}}{{{{10}^{ - 9}}{\rm{ }}C}}} \right)\\ &= 1.04{\rm{ }}nC\end{aligned}\]

Hence, the magnitude of the charges will be \[1.04{\rm{ }}nC\].

Charge will be consistent

(b)

Since, the tap becomes charged when pulled from dispenser which generates static electricity which produces electrostatic force that supports the weight of piece of tape.

Hence, the magnitude of the charge will be consistent with the static electricity as long as the tap is pulled from the dispenser.