StudySmarter AI is coming soon!

- :00Days
- :00Hours
- :00Mins
- 00Seconds

A new era for learning is coming soonSign up for free

Suggested languages for you:

Americas

Europe

78GP

Expert-verifiedFound in: Page 473

Book edition
7th

Author(s)
Douglas C. Giancoli

Pages
978 pages

ISBN
978-0321625922

**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 \[0.093\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\].

** **

**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.**

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,\[{q_2} = - 650\;{\rm{\mu C}}\].

The potential difference can be calculated as:

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

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:

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

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 \[0.093\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}\].

94% of StudySmarter users get better grades.

Sign up for free