Energy per coulomb
Energy of a pair depends on both charges. Often one set of charges is fixed and we want to know about many other charges placed near it. Then it is useful to divide out the charge that is brought in. Bring a small positive charge \(q_0\) slowly from infinity to a point P. The outside work \(W\) divided by \(q_0\) depends only on the fixed charges and on the point.
The electric potential \(V\) at a point is the work per unit charge done by an outside agent on a small positive charge. The charge is brought slowly from infinity to that point: \(V = W/q_0\). Its unit is the volt, \(1\,\text{V} = 1\,\text{J/C}\).
So potential is a property of a point in space, and it exists whether or not a charge sits there. Potential energy is a property of a system of charges. If a charge \(q\) is placed at a point of potential \(V\), the system gains the energy \(U = qV\). A charge of \(2\,\mu\text{C}\) placed where \(V = 500\,\text{V}\) gives the system \(10^{-3}\,\text{J}\).
An outside agent brings a charge of \(2\,\mu\text{C}\) slowly from infinity to a point P and does \(4 \times 10^{-3}\,\text{J}\) of work. Find the potential at P. How much work would it take to bring \(-1\,\mu\text{C}\) to P instead?
\(V = W/q = 4 \times 10^{-3}/(2 \times 10^{-6}) = 2000\,\text{V}\).
For \(-1\,\mu\text{C}\), \(W = qV = (-1 \times 10^{-6}) \times 2000 = -2 \times 10^{-3}\,\text{J}\). The work is negative: the field pulls the negative charge in, and the agent only holds it back.
Values used in this lesson: \(k = 9 \times 10^9\,\text{N m}^2/\text{C}^2\), \(e = 1.6 \times 10^{-19}\,\text{C}\), \(m_e = 9.1 \times 10^{-31}\,\text{kg}\). This example needs none of them.
Two signs, stated once
When a charge \(q\) moves from a point of potential \(V_i\) to one of potential \(V_f\), its potential energy changes by \(q(V_f - V_i)\). A slow outside agent supplies exactly that. The field does the opposite work. We use these two lines for the rest of the book.
The electric field does \(20\,\text{J}\) of work in moving a charge of \(4\,\text{C}\) from a point at \(10\,\text{V}\) to a point B. What is the potential of B, in volts?
The potential of a point charge
Divide the energy of a pair by the charge brought in. A test charge \(q_0\) at a distance \(r\) from a point charge \(q\) gives \(U = kqq_0/r\). So the potential of the point charge is \(V = U/q_0 = kq/r\). It is positive near a positive charge and negative near a negative one.
The sim shows the difference between a number and a vector. At \(20\,\text{cm}\) the potential is \(9 \times 10^9 \times 4 \times 10^{-6}/0.2 = 1.8 \times 10^5\,\text{V}\) in every direction. At \(40\,\text{cm}\) it is \(9 \times 10^4\,\text{V}\). The field there, \(kq/r^2\), has a size and a direction; the potential has only a value.
A charge of \(5\,\mu\text{C}\) sits at the origin. What is the potential at the point \((3\,\text{m}, 4\,\text{m})\)?
What is the potential \(30\,\text{cm}\) from a charge of \(-6\,\mu\text{C}\), in volts? Give the sign.
Potentials add as numbers
Work done against several charges adds up, so potentials add: \(V = \sum kq_i/r_i\). Each term is a signed number, and no direction enters. This is far easier than adding fields, which need components. But it hides a trap: a zero potential does not mean a zero field.
Charges of \(+2\,\mu\text{C}\) and \(-2\,\mu\text{C}\) are \(20\,\text{cm}\) apart. At the midpoint P the potential is zero. What is the electric field at P?
Now choose the second chip. Four charges of \(2\,\mu\text{C}\) sit on a square whose diagonal is \(20\,\text{cm}\), so each is \(10\,\text{cm}\) from the centre. The potential there is \(4 \times kq/r = 7.2 \times 10^5\,\text{V}\). The fields of opposite corners cancel in pairs, so \(E = 0\). Potential and field are independent pieces of information about a point.
Charges of \(+3\,\mu\text{C}\) and \(-2\,\mu\text{C}\) are \(30\,\text{cm}\) apart. Find the two points on the line through them where the potential is zero.
Measure \(x\) from the \(+3\,\mu\text{C}\) towards the \(-2\,\mu\text{C}\). The potential is zero where \(3/x = 2/(\text{distance from the } -2\,\mu\text{C})\).
Between the charges: \(3/x = 2/(0.3 - x)\). Outside, beyond the smaller charge: \(3/x = 2/(x - 0.3)\).
Between: \(0.9 - 3x = 2x\), so \(x = 0.18\,\text{m}\). Outside: \(3x - 0.9 = 2x\), so \(x = 0.9\,\text{m}\).
The potential is zero \(18\,\text{cm}\) and \(90\,\text{cm}\) from the \(+3\,\mu\text{C}\), as the third chip of the figure above shows. The field is not zero at either point.
Charges of \(100\,\mu\text{C}\), \(-50\,\mu\text{C}\), \(20\,\mu\text{C}\) and \(-60\,\mu\text{C}\) sit at the corners of a square of side \(\sqrt2\,\text{m}\). What is the potential at the centre, in volts?
Potential difference
Only differences of potential enter a work, so the potential difference \(V_B - V_A\) is what batteries, meters and experiments deal with. Move \(2\,\mu\text{C}\) slowly from a point at \(40\,\text{V}\) to one at \(100\,\text{V}\). The agent does \(W = 2 \times 10^{-6} \times 60 = 1.2 \times 10^{-4}\,\text{J}\), and the field does \(-1.2 \times 10^{-4}\,\text{J}\).
Left free, a charge moves the other way. A positive charge released at \(200\,\text{V}\) is pushed to lower potential. If it reaches \(50\,\text{V}\), its potential energy changes by \(q(V_f - V_i) = 2 \times 10^{-6} \times (-150)\) \(= -3 \times 10^{-4}\,\text{J}\). Energy is conserved, so it gains \(3 \times 10^{-4}\,\text{J}\) of kinetic energy.
The electron volt
For single electrons and ions the joule is far too large. An electron crossing a potential difference of one volt gains \(eV = 1.6 \times 10^{-19}\,\text{J}\). This amount is the electron volt, written eV. A particle of charge \(e\) crossing \(100\,\text{V}\) gains \(100\,\text{eV}\), and an alpha particle of charge \(2e\) crossing \(1000\,\text{V}\) gains \(2\,\text{keV}\).
An electron starts from rest and is accelerated through \(100\,\text{V}\). Find its kinetic energy in joules and in electron volts, and its speed.
\(K = eV = 1.6 \times 10^{-19} \times 100 = 1.6 \times 10^{-17}\,\text{J}\), which is \(100\,\text{eV}\).
From \(\tfrac12 m_e v^2 = K\): \(v = \sqrt{2 \times 1.6 \times 10^{-17}/9.1 \times 10^{-31}} = 5.9 \times 10^6\,\text{m/s}\).
That is about 2 % of the speed of light, reached across a gap of any length. Only the potential difference matters.
In a hydrogen atom the electron is \(5.3 \times 10^{-11}\,\text{m}\) from the proton. The potential energy of the pair is \(-4.35 \times 10^{-18}\,\text{J}\); ignore the electron's kinetic energy. What is this energy in electron volts?
A charge of \(-3\,\mu\text{C}\) is moved slowly from a point at \(0\,\text{V}\) to a point at \(+100\,\text{V}\). Which statements are true?
Potential is like height and charge like mass. A positive charge rolls downhill, to lower potential, as a ball rolls to lower height. A negative charge is like a helium balloon: it rises, towards higher potential.
In 1800 the Italian physicist Alessandro Volta stacked discs of zinc and copper with cloth soaked in brine between them. The pile gave a steady push to charge, the first battery. For the first time a steady potential difference could be kept up for hours. When a unit was named for the work per coulomb, it was named after him: the volt.
Crows sit calmly on the bare 11 kV lines that carry power along Indian roads. The wire is thousands of volts above the ground, but both feet of the bird touch the same wire. The potential difference between the feet is almost zero, so no current flows through the bird. A crow that touched the wire and a pole at once would bridge a large potential difference, and that is dangerous.
- Treating a zero potential as a zero field, or a zero field as a zero potential.Potential adds as numbers and field as vectors; find each separately.
- Adding potentials with components or directions.Add the signed values \(kq_i/r_i\); potential has no direction.
- Using \(kq/r^2\) for the potential.\(V = kq/r\); \(kq/r^2\) is the size of the field.
- Forgetting the sign of a negative charge in \(W = q\,\Delta V\).Put \(q\) in with its sign every time.