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