A charge of +4.0 µC and a charge of -1.0 µC are fixed 0.30 m apart in vacuum, as the figure shows…

A charge of $+4.0\ \mu\text{C}$ and a charge of $-1.0\ \mu\text{C}$ are fixed $0.30\ \text{m}$ apart in vacuum, as the figure shows. The distance $x$ is measured along their line, from the positive charge towards the negative one. Take $\dfrac{1}{4\pi\varepsilon_0} = 9.0 \times 10^{9}\ \text{N m}^2\text{C}^{-2}$ and take the potential to be zero very far away.

A horizontal straight line with a distance scale x in metres, measured from the left charge. A charge of plus 4.0 microcoulombs sits at x equal to zero and a charge of minus 1.0 microcoulomb sits at x equal to 0.30 m. The separation of 0.30 m is marked with a double arrow. The line continues to the right beyond the second charge and a little to the left of the first, ending in arrows. No other point is marked.

The potential is zero at two points of this line. The electric field is zero at one point of this line.

What is the potential at the midpoint of the two charges?
At what distance from the $+4.0\ \mu\text{C}$ charge is the potential zero, at the point between the charges?
At what distance from the $+4.0\ \mu\text{C}$ charge is the potential zero, at the other point?
At what distance from the $+4.0\ \mu\text{C}$ charge is the electric field zero?
How much work must an external agent do to bring a charge of $+2.0\ \mu\text{C}$ slowly from very far away to the point where the field is zero?

Do not assume that the work is zero just because the field is zero there.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Given. $q_1 = +4.0\ \mu\text{C}$ at $x = 0$ , $q_2 = -1.0\ \mu\text{C}$ at $x = 0.30\ \text{m}$ .

(a) Midpoint. Each charge is $0.15\ \text{m}$ away: $V = \dfrac{k(q_1 + q_2)}{0.15\ \text{m}} = \dfrac{9.0\times10^{9} \times 3.0\times10^{-6}}{0.15} = 180\ \text{kV}$ .

(b) Zero between the charges. $\dfrac{4.0}{x} = \dfrac{1.0}{0.30 - x}$ gives $4(0.30 - x) = x$ , so $x = 0.24\ \text{m}$ .

(c) The other zero. Beyond the negative charge, $\dfrac{4.0}{x} = \dfrac{1.0}{x - 0.30}$ gives $4(x - 0.30) = x$ , so $x = 0.40\ \text{m}$ .

(d) Zero field. Beyond the negative charge, the two fields point in opposite directions. $\dfrac{4.0}{x^2} = \dfrac{1.0}{(x - 0.30)^2}$ gives $2(x - 0.30) = x$ , so $x = 0.60\ \text{m}$ .

(e) Work. At this point, $V = k\left(\dfrac{4.0\times10^{-6}}{0.60} - \dfrac{1.0\times10^{-6}}{0.30}\right) = 30.0\ \text{kV}$ , not zero. The work is $W = q_tV = 2.0\times10^{-6}\ \text{C} \times 30000\ \text{V} = 0.060\ \text{J}$ .

Answer: (a) $180\ \text{kV}$ . (b) $0.24\ \text{m}$ . (c) $0.40\ \text{m}$ . (d) $0.60\ \text{m}$ . (e) $0.060\ \text{J}$ .

Check. Put each distance back. At $0.24\ \text{m}$ , $\dfrac{4.0}{0.24} - \dfrac{1.0}{0.06} = 0$ . At $0.40\ \text{m}$ , $\dfrac{4.0}{0.40} - \dfrac{1.0}{0.10} = 0$ . For the field at $0.60\ \text{m}$ , $\dfrac{4.0}{0.36} = \dfrac{1.0}{0.09}$ .

Watch out. The two zeros of the potential are not the zero of the field. If you set the distances in the ratio $4:1$ for the field, you get $0.40\ \text{m}$ instead of $0.60\ \text{m}$ : the field goes as $1/x^2$ , so the ratio of the distances is $2:1$ .

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A charge of +4.0 µC and a charge of -1.0 µC are fixed 0.30 m apart in vacuum, as the figure shows…

The worked solution

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