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Found in: Page 473

### Physics Principles with Applications

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

See the step by step solution

## 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 $0.093\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.} {\rm{s}}}$.