The balloon in Fig. 16–48 was rubbed on a student’s hair. Explain why the water drip curves instead of falling vertically.

Short Answer

Expert verified

The water drip curves instead of falling vertically due to the electrostatic force of attraction between the charged balloon and the charge of opposite polarity induced on the stream of water.

Step by step solution

01

Understanding charging by induction

When a charged object induces a charge on another uncharged object without actually coming in contact with it, the process is termed as charging by induction. The charges developed on the uncharged body are known as induced charges.

02

Explanation for curving of water drip

When the balloon is rubbed on the student’s hair, it gets charged due to rubbing. When this charged balloon is brought near the water drip, the charges get induced in the water stream. Consequently, the charges of the water that has a polarity opposite to that on the balloon start moving to the face of the water stream, which is closer to the balloon due to the electrostatic force of attraction between the charges.

The charges that are accumulated on the closer face of the water stream cannot move out of it. As a result of the accumulation of charges of opposite polarity on the closer face of the water stream, it gets attracted toward the balloon. It thus follows a curved path instead of falling vertically.

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Most popular questions from this chapter

(II) Determine the magnitude of the acceleration experienced by an electron in an electric field of\({\bf{756 N/C}}\).How does the direction of the acceleration depend on the direction of the field at that point?

We are usually not aware of the electric force acting between two everyday objects because

(a) the electric force is one of the weakest forces in nature.

(b) the electric force is due to microscopic-sized particles such as electrons and protons.

(c) the electric force is invisible.

(d) most everyday objects have as many plus charges as minus charges.

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?

We wish to determine the electric field at a point near a positively charged metal sphere (a good conductor). We do so by bringing a small positive test charge, \({q_{\rm{o}}}\), to this point and measure the force \({F_{\rm{o}}}\) on it. \({{{F_{\rm{o}}}} \mathord{\left/{\vphantom {{{F_{\rm{o}}}} {{q_{\rm{o}}}}}} \right. \\{{q_{\rm{o}}}}}\) will be ____________ the electric field \({\bf{\vec E}}\) as it was at that point before the test charge was present.

(a) greater than

(b) less than

(c) equal to

A point charge of mass 0.185 kg, and net charge\( + {\bf{0}}{\bf{.340 }}\mu {\bf{C}}\)hangs at rest at the end of an insulating cord above a large sheet of charge. The horizontal sheet of fixed uniform charge creates a uniform vertical electric field in the vicinity of the point charge. The tension in the cord is measured to be 5.18 N. Calculate the magnitude and direction of the electric field due to the sheet of charge (Fig. 16–67).

FIGURE 16–67 Problem 61.

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