(a) How strong is the attractive force between a glass rod with a 0.700μCcharge and a silk cloth with a -0.600μCcharge, which are 12.0 cm apart, using the approximation that they act like point charges? (b) Discuss how the answer to this problem might be affected if the charges are distributed over some area and do not act like point charges.

Short Answer

Expert verified

(a) The force of attraction between the glass rod and the silk cloth is 0.262 N. (b) If the charges are distributed over some area, the thing would get complicated.

Step by step solution

01

Given Data

  • Charge on the glass rodq1=0.700μC
  • Charge on silk clothq2=-0.600μC
  • Separation between rod and cloth=12.0cm
02

Coulomb’s force

When two-point charges are separated by some distance, the Coulomb force is given as,

F=Kq1q2r2

Here, K is the electrostatic force constant, q1and q2are the magnitude of charges, and ris the separation between the charges.

03

Attractive force between glass rod and silk cloth

The attractive force between the glass rod and silk cloth is,

F=Kq1q2r2

Here, K is the electrostatic force constant K=9×109N-m2/C2, q1is the magnitude of charge on glass rod q1=0.700μC, q2is the magnitude of charge on the silk cloth q2=0.600μC and r is the separation between the glass rod and silk cloth (r = 12.0 cm) .

Substituting all known values,

F=9×109N-m2/C2×0.700μC×0.600μC12.0cm2=9×109N-m2/C2×0.700μC×10-6C1μC×0.600μC×10-6C1μC12.0cm×1m100cm-2=0.262N

Hence, the force of attraction between the glass rod and the silk cloth is 0.262 N .

04

When charges are distributed over some area

If the charges are distributed over some area, the resultant force between the glass rod and the silk cloth will be the vector sum of each force on each pair of charge. Hence, the things would get complicated.

Thus the glass rod and silk cloth will experience a force of 0.262 N between them.

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Most popular questions from this chapter

Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated protons separated by\({\bf{2}}.{\bf{00}}{\rm{ }}{\bf{nm}}\)(a typical distance between gas atoms). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.

Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance between the centers of the balls. Construct a problem in which you calculate the electric field (magnitude and direction) due to the balls at various points along a line running through the centers of the balls and extending to infinity on either side. Choose interesting points and comment on the meaning of the field at those points. For example, at what points might the field be just that due to one ball and where does the field become negligibly small? Among the things to be considered are the magnitudes of the charges and the distance between the centers of the balls. Your instructor may wish for you to consider the electric field off axis or for a more complex array of charges, such as those in a water molecule.

Sketch the electric field lines in the vicinity of the charged insulator in Figure 18.51 noting its nonuniform charge distribution.

Figure 18.51

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Using Figure, explain, in terms of Coulomb’s law, why a polar molecule (such as in Figure 18.43) is attracted by both positive and negative charges.

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