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Physics Principles with Applications
Found in: Page 473
Physics Principles with Applications

Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922

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Short Answer

Question: Near the surface of the Earth there is an electric field of about \({\bf{150}}\;{{\bf{V}} \mathord{\left/{\vphantom {{\bf{V}} {\bf{m}}}} \right.} {\bf{m}}}\)which points downward. Two identical balls with mass \({\bf{m = 0}}{\bf{.670}}\;{\bf{kg}}\) are dropped from a height of 2.00 m, but one of the balls is positively charged with \({{\bf{q}}_{\bf{1}}}{\bf{ = 650}}\;{\bf{\mu C}}\), and the second is negatively charged with \({{\bf{q}}_{\bf{2}}}{\bf{ = }} - {\bf{650}}\;{\bf{\mu C}}\). Use conservation of energy to determine the difference in the speed of the two balls when they hit the ground. (Neglect air resistance.)

The difference in the speed of the balls when they are hitting the ground is \[0.093\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\].

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1:Understanding of the electric potential difference  

The value of the electric potential difference can be calculated by multiplying the value of the magnitude of the electric field with the distance between the two points.

Step 2: Given information 

The electric field is,\[E = 150\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} {\rm{m}}}} \right.} {\rm{m}}}\].

The mass of the balls is,\[m = 0.670\;{\rm{kg}}\].

The height is,\[h = 2.00\;{\rm{m}}\].

The charge on one ball is,\[{q_1} = 650\;{\rm{\mu C}}\].

The charge on another ball is,\[{q_2} = - 650\;{\rm{\mu C}}\].

Step 3: Evaluation of the potential difference 

The potential difference can be calculated as:

\[\begin{array}{c}{V_1} - {V_2} = Eh\\ = \left({150\;{{\rm{V}} \mathord{\left/{\vphantom {{\rm{V}} {\rm{m}}}} \right.} {\rm{m}}}} \right)\left( {2\;{\rm{m}}} \right)\\ = 300\;{\rm{V}}\end{array}\]

Step 4: Evaluation of the speed of both the balls 

Apply the conservation of energy principle at the time of dropping of the ball and the ball at the surface of the Earth.

\[\begin{array}{c}mgh + q{V_1} = \frac{1}{2}m{v^2} + q{V_2}\\\frac{1}{2}m{v^2} = mgh + q\left( {{V_1} - {V_2}} \right)\\v = \sqrt {\frac{{2mgh + 2q\left( {{V_1} - {V_2}} \right)}}{m}} \\v = \sqrt {2gh + \frac{{2q\left( {{V_1} - {V_2}} \right)}}{m}} \end{array}\]

The velocity of the positively charged ball can be calculated as:

\[\begin{array}{c}{v_{{q_1}}} = \sqrt {2gh + \frac{{2{q_1}\left( {{V_1} - {V_2}} \right)}}{m}} \\ = \sqrt {2\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {2\;{\rm{m}}} \right) + \frac{{2\left\{ {\left( {{\rm{650}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} \right\}\left( {300\;{\rm{V}}} \right)}}{{\left( {0.670\;{\rm{kg}}} \right)}}} \\ = 6.307\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{array}\]

The velocity of the negatively charged ball can be calculated as:

\[\begin{array}{c}{v_{{q_2}}} = \sqrt {2gh + \frac{{2{q_2}\left( {{V_1} - {V_2}} \right)}}{m}} \\ = \sqrt {2\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right)\left( {2\;{\rm{m}}} \right) + \frac{{2\left\{ {\left( {{\rm{ - 650}}\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} \right\}\left( {300\;{\rm{V}}} \right)}}{{\left( {0.670\;{\rm{kg}}} \right)}}} \\ = 6.214\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{array}\]

Step 5: Evaluation of the difference in the speed of the balls when they are hitting the ground 

The difference in the speed of the balls when they are hitting the ground can be calculated as:

\[\begin{array}{c}v = {v_{{q_1}}} - {v_{{q_2}}}\\ = \left( {6.307\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right) - \left( {6.214\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}} \right)\\ = 0.093\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\end{array}\]

Thus, the difference in the speed of the balls when they are hitting the ground is \[0.093\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\].

Most popular questions for Physics Textbooks

Question66:(III) In a given CRT, electrons are accelerated horizontally by 9.0 kV. They then pass through a uniform electric field E for a distance of 2.8 cm, which deflects them upward so they travel 22 cm to the top of the screen, 11 cm above the center. Estimate the value of E.

Can a particle ever move from a region of low electric potential to one of high potential and yet have its electric potential energy decrease? Explain.

Question: An electron is accelerated horizontally from rest by a potential difference of 2200 V. It then passes between two horizontal plates 6.5 cm long and 1.3 cm apart that have a potential difference of 250 V (Fig. 17–50). At what angle \(\theta \) will the electron be traveling after it passes between the plates?

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A dielectric is pulled out from between the plates of a capacitor which remains connected to a battery. What changes occur to (a) the capacitance, (b) the charge on the plates, (c) the potential difference, (d) the energy stored in the capacitor, and (e) the electric field? Explain your answers.
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