Question

Question: A \({\bf{3}}{\bf{.4}}\;{\bf{\mu C}}\) and a \({\bf{ - 2}}{\bf{.6}}\;{\bf{\mu C}}\) charge are placed 2.5 cm apart. At what points along the line joining them is (a) the electric field zero, and (b) the electric potential zero?

Short answer

(a)The electric field is zero at 17.4 cm from the negative charge and on the opposite side from the positive charge.

(b) The one point where the net electric potential is zero is at 1.1 cm from the negative charge and towards the positive charge. The second point where the net electric potential is zero is at 8.1 cm from the negative charge and away from the positive charge.

Step-by-step solution

Understanding of electric field  

The magnitude of the electric field can be calculated by evaluating the charge's value and the charge's distance from a particular point.

In this question, the direction of the electric field at a point is away from the positive charge and towards the negative charge.

Given information 

The first charge is,\( {q_1} = 3.4\;{\rm{\mu C}}\).

The second charge is,\( {q_2} = - 2.6\;{\rm{\mu C}}\).

The separation between the charges is,\( d = 2.5\;{\rm{cm}}\).

(a) Evaluation of the point where the net electric field is zero. 

The schematic diagram for the problem can be drawn as:

Here,\({E_1}\)and \({E_2}\)are the electric fields due to charge\({q_1}\)and \({q_2}\) respectively and \(P\) is the point at a distance \(x\) from the negative charge along the line joining the two charges where the net electric field is zero.

For the net electric field to be zero, we can write:

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

Solve further the above equation.

\( \begin{aligned}{c}\frac{{d + x}}{x} = \sqrt {\frac{{{q_1}}}{{{q_2}}}} \\x = \frac{d}{{\sqrt {\left| {\frac{{{q_1}}}{{{q_2}}}} \right| - 1} }}\\ = \frac{{\left( {2.5\;{\rm{cm}}} \right)}}{{\sqrt {\left| {\frac{{\left( {3.4\;{\rm{\mu C}}} \right)}}{{\left( { - 2.6\;{\rm{\mu C}}} \right)}}} \right| - 1} }}\\ = 17.4\;{\rm{cm}}\end{aligned}\)

Thus, the electric field is zero at 17.4 cm from the negative charge and on the opposite side from the positive charge.

(b) Evaluation of the points where the net electric potential is zero.

The schematic diagram can be drawn as:

The\(P\) and \(R\) are the two points where the electric potential can be zero. The points where the electric potential is zero lies near the smaller magnitude charge.

For the net electric potential to be zero at point R, one can write:

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

Solve further the above equation.

\( \begin{aligned}{c}{x_1} = \frac{{ - {q_2}d}}{{\left( {{q_1} - {q_2}} \right)}}\\ = \frac{{ - \left( { - 2.6\;{\rm{\mu C}}} \right)\left( {2.5\;{\rm{cm}}} \right)}}{{\left( {\left( {3.4\;{\rm{\mu C}}} \right) - \left( { - 2.6\;{\rm{\mu C}}} \right)} \right)}}\\ = 1.08\;{\rm{cm}}\\ \approx 1.1\;{\rm{cm}}\end{aligned}\)

Thus, the one point where the net electric potential is zero is at 1.1 cm from the negative charge and towards the positive charge.

For the net electric potential to be zero at point P, one can write:

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

Solve further the above equation.

\( \begin{aligned}{c}{x_2} = - \frac{{{q_2}d}}{{\left( {{q_1} + {q_2}} \right)}}\\ = \frac{{ - \left( { - 2.6\;{\rm{\mu C}}} \right)\left( {2.5\;{\rm{cm}}} \right)}}{{\left( {\left( {3.4\;{\rm{\mu C}}} \right) + \left( { - 2.6\;{\rm{\mu C}}} \right)} \right)}}\\ = 8.12\;{\rm{cm}} \approx {\rm{8}}{\rm{.1}}\;{\rm{cm}}\end{aligned}\)

Thus, the second point where the net electric potential is zero is at 8.1 cm from the negative charge and away from the positive charge.