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Chapter 23: Problem 37

A spherical water drop \(50.0 \mu \mathrm{m}\) in diameter has a uniformly distributed charge of \(+20.0 \mathrm{pC}\). Find (a) the potential at its surface and (b) the potential at its center.

Short Answer

Expert verified
Answer: The potential at both the surface and the center of the water drop is 7192 volts.

Step by step solution

01

Identify given values

We are given the diameter (\(50.0 \mu m\)) of the water drop, and its charge (\(+20.0 pC\)). Let's convert these values to meters and Columbs respectively: Diameter = \(50.0 \mu m = 50.0 × 10^{-6} m\) Charge = \(+20.0 pC = 20.0 × 10^{-12} C\)
02

Find the radius of the water drop

To find the radius of the water drop, divide the diameter in meters by 2: Radius = \(\frac{50.0 × 10^{-6} m}{2} = 25.0 × 10^{-6} m\)
03

Calculate potential at the surface

Use the formula for potential to find the potential at the surface of the charged sphere: \(V_{surface} = \frac{kQ}{r}\) \(V_{surface} = \frac{8.99 × 10^9 Nm^2 C^{-2} × 20.0 × 10^{-12} C}{25.0 × 10^{-6} m}\) \(V_{surface} = 7192 V\) The potential at the surface of the water drop is 7192 volts.
04

Calculate potential at the center

For a uniformly charged sphere, the potential at the center is equal to the potential at the surface (since the electric field inside is zero). So the potential at the center is: \(V_{center} = V_{surface}\) \(V_{center} = 7192 V\) The potential at the center of the water drop is also 7192 volts.

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Most popular questions from this chapter

Suppose that an electron inside a cathode ray tube starts from rest and is accelerated by the tube's voltage of \(21.9 \mathrm{kV}\). What is the speed (in \(\mathrm{km} / \mathrm{s}\) ) with which the electron (mass \(=9.11 \cdot 10^{-31} \mathrm{~kg}\) ) hits the screen of the tube?

A particle with a charge of \(+5.0 \mu C\) is released from rest at a point on the \(x\) -axis, where \(x=0.10 \mathrm{~m}\). It begins to move as a result of the presence of a \(+9.0-\mu C\) charge that remains fixed at the origin. What is the kinetic energy of the particle at the instant it passes the point \(x=0.20 \mathrm{~m} ?\)

In which situation is the electric potential the highest? a) at a point \(1 \mathrm{~m}\) from a point charge of \(1 \mathrm{C}\) b) at a point \(1 \mathrm{~m}\) from the center of a uniformly charged spherical shell of radius \(0.5 \mathrm{~m}\) with a total charge of \(1 \mathrm{C}\) c) at a point \(1 \mathrm{~m}\) from the center of a uniformly charged rod of length \(1 \mathrm{~m}\) and with a total charge of \(1 \mathrm{C}\) d) at a point \(2 \mathrm{~m}\) from a point charge of \(2 \mathrm{C}\) e) at a point \(0.5 \mathrm{~m}\) from a point charge of \(0.5 \mathrm{C}\)

Each of the following pairs of charges are separated by a distance \(d\). Which pair has the highest potential energy? a) \(+5 \mathrm{C}\) and \(+3 \mathrm{C}\) d) \(-5 \mathrm{C}\) and \(+3 \mathrm{C}\) b) \(+5 \mathrm{C}\) and \(-3 \mathrm{C}\) e) All pairs have the \(\begin{array}{ll}\text { c) }-5 \mathrm{C} \text { and }-3 \mathrm{C} & \text { same potential energy. }\end{array}\)

A negatively charged particle revolves in a clockwise direction around a positively charged sphere. The work done on the negatively charged particle by the electric field of the sphere is a) positive. b) negative. c) zero.
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