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(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

Expert verified

(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

01

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.

02

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 \[1.04{\rm{ }}nC\].

03

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.

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Most popular questions from this chapter

(a) Using the symmetry of the arrangement, determine the direction of the electric field at the center of the square in Figure 18.53, given that\({q_a} = {q_b} = - {\rm{1}}{\rm{.00 }}\mu {\rm{C}}\)and\({q_c} = {q_d} = + {\rm{1}}{\rm{.00 mC}}\). (b) Calculate the magnitude of the electric field at the location of\(q\), given that the square is\(5.00{\rm{ cm}}\)on a side.

Figure 18.43 shows the charge distribution in a water molecule, which is called a polar molecule because it has an inherent separation of charge. Given water’s polar character, explain what effect humidity has on removing excess charge from objects.

Figure 18.43 Schematic representation of the outer electron cloud of a neutral water molecule. The electrons spend more time near the oxygen than the hydrogens, giving a permanent charge separation as shown. Water is thus a polar molecule. It is more easily affected by electrostatic forces than molecules with uniform charge distributions.

A\(5.00{\rm{ g}}\)charged insulating ball hangs on a\(30.0{\rm{ cm}}\)long string in a uniform horizontal electric field as shown in Figure 18.56. Given the charge on the ball is\(1.00{\rm{ }}\mu {\rm{C}}\), find the strength of the field.

Figure 18.56 A horizontal electric field causes the charged ball to hang at an angle of\(8.00^\circ \).

Common static electricity involves charges ranging from nanocoulombs to microcoulombs. (a) How many electrons are needed to form a charge of –2.00 nC (b) How many electrons must be removed from a neutral object to leave a net charge of 0.500 µC?

What is the force on the charge located at \(x = 8.00{\rm{ }}cm\) in Figure 18.52(a) given that \(q = 1.00{\rm{ }}\mu C\)?

Figure 18.52 (a) Point charges located at \[{\bf{3}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{11}}.{\bf{0}}{\rm{ }}{\bf{cm}}\] along the x-axis. (b) Point charges located at \[{\bf{1}}.{\bf{00}},{\rm{ }}{\bf{5}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{14}}.{\bf{0}}{\rm{ }}{\bf{cm}}\] along the x-axis

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