(II) Two point charges, \({\bf{3}}{\bf{.0}}\;{\bf{\mu C}}\) and \({\bf{ - 2}}{\bf{.0}}\;{\bf{\mu C}}\) are placed 4.0 cm apart on the x axis. At what points along the x axis is (a) the electric field zero and (b) the potential zero?

The electric potential may be defined as the potential energy per charge. It is a property of a location within an electric field. Its value is the same for all charges at a particular location.

The first charge is \({q_1} = 3.0\;{\rm{\mu C}}\).

The second charge is \({q_2} = - 2.0\;{\rm{\mu C}}\).

The separation between the charges is \(d = 4.0\;{\rm{cm}}\).

The following is the figure showing two point charges separated by a distance.

Let, P be the point where the net field due to charges \({q_1}\) and \({q_2}\) will be zero. The field will be zero, when it is closer to the weaker charge, say \({q_2}\). In between the two charges the electric field is parallel to each other and cannot cancel.

The field at point P due to charge \({q_1}\) is as follows:

\({E_1} = \frac{{k{q_1}}}{{{{\left( {x + d} \right)}^2}}}\)

The field at point P due to charge \({q_2}\) is as follows:

\({E_2} = \frac{{k{q_2}}}{{{{\left( x \right)}^2}}}\)

The expression for the net electric field can be written as:

\(\begin{aligned}{E_{{\rm{net}}}} &= {E_1} - {E_2} = 0\\{E_1} - {E_2} &= 0\\\left( {\frac{{k{q_1}}}{{{{\left( {x + d} \right)}^2}}}} \right) - \left( {\frac{{k{q_2}}}{{{{\left( x \right)}^2}}}} \right) &= 0\\\left( {\frac{{{q_1}}}{{{{\left( {x + d} \right)}^2}}}} \right) &= \left( {\frac{{{q_2}}}{{{x^2}}}} \right)\end{aligned}\)

Solve further as,

\(\begin{aligned}\sqrt {\frac{{{q_1}}}{{{q_2}}}} &= \frac{{\left( {x + d} \right)}}{{\left( x \right)}}\\x &= \left( {x + d} \right)\sqrt {\frac{{{q_1}}}{{{q_2}}}} \\x &= x\sqrt {\frac{{{q_1}}}{{{q_2}}}} + d\sqrt {\frac{{{q_1}}}{{{q_2}}}} \\x\left( {1 - \sqrt {\frac{{{q_1}}}{{{q_2}}}} } \right) &= d\sqrt {\frac{{{q_1}}}{{{q_2}}}} \\x &= \frac{{d\sqrt {{q_2}} }}{{\sqrt {{q_1}} - \sqrt {{q_2}} }}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}x &= \frac{{\left( {{\rm{4}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)\sqrt {\left( {\left| { - 2.0\;{\rm{\mu C}}} \right|} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} }}{{\sqrt {\left( {\left| {3.0\;{\rm{\mu C}}} \right|} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} - \sqrt {\left( {\left| { - 2.0\;{\rm{\mu C}}} \right|} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} }}\\x &= \left( {0.18\;{\rm{m}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{2}}}\;{\rm{cm}}}}{{{\rm{1}}\;{\rm{m}}}}} \right)\\x &= 18.0\;{\rm{cm}}\end{aligned}\)

Thus, the electric field is zero at a distance of \(18.0\;{\rm{cm}}\) from the weaker charge.

The following is the figure showing two point charges separated by a distance.

The potential is zero when it is closer to the weaker charge. Let, \({x_1}\) be the point between the two charges, and \({x_2}\) be the point to the left of weaker charge where the electric potential is zero.

The value of the \({x_1}\) can be calculated as:

\(\begin{aligned}{V_1} &= \frac{{k{q_1}}}{{\left( {d - {x_1}} \right)}} + \frac{{k{q_2}}}{{{x_1}}} = 0\\\frac{{k{q_1}}}{{\left( {d - {x_1}} \right)}} &= - \frac{{k{q_2}}}{{{x_1}}}\\{q_1}{x_1} = - {q_2}\left( {d - {x_1}} \right)\\\left( {{q_1} - {q_2}} \right){x_1} &= - {q_2}d\\{x_1} &= \frac{{ - {q_2}d}}{{\left( {{q_1} - {q_2}} \right)}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{x_1} &= \frac{{ - \left( {\left( { - 2.0\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} \right)\left( {\left( {{\rm{4}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right)}}{{\left( {\left( {3.0\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} \right) - \left( {\left( { - 2.0\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} \right)}}\\{x_1} &= \left( {0.016\;{\rm{m}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{2}}}\;{\rm{cm}}}}{{{\rm{1}}\;{\rm{m}}}}} \right)\\{x_1} &= 1.6\;{\rm{cm}}\end{aligned}\)

Thus, the electric potential is zero at a distance of \(1.6\;{\rm{cm}}\) from the weaker charge between the two charges.

The value of the \({x_2}\) can be calculated as:

\(\begin{aligned}{V_2} &= \frac{{k{q_1}}}{{d + {x_2}}} + \frac{{k{q_2}}}{{{x_2}}} &= 0\\\frac{{k{q_1}}}{{d + {x_2}}} &= - \frac{{k{q_2}}}{{{x_2}}}\\{q_1}{x_2} &= {q_2}\left( {d + {x_2}} \right)\\\left( {{q_1} + {q_2}} \right){x_2} &= - {q_2}d\\{x_2} &= \frac{{ - {q_2}d}}{{\left( {{q_1} + {q_2}} \right)}}\end{aligned}\)

Substitute the values in the above equation.

\(\begin{aligned}{x_2} &= \frac{{ - \left( {\left( { - 2.0\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} \right)\left( {\left( {{\rm{4}}\;{\rm{cm}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 2}}}}\;{\rm{m}}}}{{{\rm{1}}\;{\rm{cm}}}}} \right)} \right)}}{{\left( {\left( {3.0\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} \right) + \left( {\left( { - 2.0\;{\rm{\mu C}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{{\rm{ - 6}}}}\;{\rm{C}}}}{{{\rm{1}}\;{\rm{\mu C}}}}} \right)} \right)}}\\{x_2} &= \left( {0.08\;{\rm{m}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{2}}}\;{\rm{cm}}}}{{{\rm{1}}\;{\rm{m}}}}} \right)\\{x_2} &= 8.0\;{\rm{cm}}\end{aligned}\)

Thus, the electric potential is zero at a distance of \(8.0\;{\rm{cm}}\) from the weaker charge away from the positive charge.