Rub a plastic rod with wool and touch it to two light pith balls. Now hold the balls close together with a pair of thin plastic sticks and let go. The balls fly apart. Nobody pushed them at the moment of release, yet they gain speed. The energy for that speed was stored somewhere while you held them.
A conservative force stores energy
In the chapter on Potential Energy and Conservation of Energy you met a rule. A force whose work does not depend on the path has a potential energy. A central force, one that points along the line to a fixed centre and depends only on the distance, always has this property. The Coulomb force between two point charges is such a force.
The previous chapter gave the size of the force. Two point charges \(q_1\) and \(q_2\) a distance \(r\) apart push or pull each other with \(F = kq_1q_2/r^2\), where \(k = 1/4\pi\varepsilon_0 = 9 \times 10^9\,\text{N m}^2/\text{C}^2\). Because the force is conservative, the pair stores electric potential energy. A change in it is minus the work done by the electric force, \(\Delta U = -W\).
Bringing a charge in from far away
Keep \(q_1\) fixed at the origin and bring \(q_2\) in along the \(x\) axis from very far away, which we call infinity. Far away the force is zero, so the natural zero of potential energy is at infinite separation. The electric force changes with \(x\), so its work is an integral.
\[ F_x = \dfrac{kq_1q_2}{x^2} \]
Why. Coulomb's law along the line; a positive \(F_x\) points away from \(q_1\), towards larger \(x\).
\[ W_\text{field} = \displaystyle\int_\infty^r \dfrac{kq_1q_2}{x^2}\,dx = -\dfrac{kq_1q_2}{r} \]
Why. the work of a varying force is the integral along the path; the integral of \(1/x^2\) is \(-1/x\), which is zero at infinity.
\[ U(r) - U(\infty) = -W_\text{field} = \dfrac{kq_1q_2}{r} \]
Why. the definition of potential energy, \(\Delta U = -W\).
\[ U(r) = \dfrac{kq_1q_2}{r} \]
Read it. the zero chosen at infinity, \(U(\infty) = 0\).
What the sign says
The charges go into the formula with their signs, so \(U\) can be positive or negative. For two like charges \(U > 0\). Outside work was needed to push them together against the repulsion, and released they would fly apart. For two unlike charges \(U < 0\). They pull together, and outside work is needed to separate them to infinity.
Read the sim as an energy diagram. A like pair rolls down its hill by moving apart, so repulsion lowers \(U\). An unlike pair sits in a well, and pulling it out towards large \(r\) raises \(U\) to zero. In both cases \(U\) changes as \(1/r\), so doubling the separation halves \(U\).
Charges of \(2\,\mu\text{C}\) and \(3\,\mu\text{C}\) are \(30\,\text{cm}\) apart. Find the potential energy of the pair. Then find it with \(-3\,\mu\text{C}\) in place of \(3\,\mu\text{C}\).
\(q_1 = 2 \times 10^{-6}\,\text{C}\), \(q_2 = 3 \times 10^{-6}\,\text{C}\), \(r = 0.3\,\text{m}\). Route: point charges, so use \(k\).
\(U = kq_1q_2/r\), with the signs of the charges put in.
\(U = 9 \times 10^9 \times 2 \times 10^{-6} \times 3 \times 10^{-6}/0.3 = 0.054/0.3 = 0.18\,\text{J}\). With \(-3\,\mu\text{C}\), \(U = -0.18\,\text{J}\).
The like pair stores \(+0.18\,\text{J}\), which would appear as kinetic energy if both were released. The unlike pair needs \(0.18\,\text{J}\) of outside work to be pulled apart to infinity.
A proton and an electron are some distance apart. What does the sign of their potential energy tell you?
Two charges of \(1\,\mu\text{C}\) each are \(9\,\text{cm}\) apart. What is the potential energy of the pair, in joules?
Work to move charges slowly
Suppose an outside agent moves a charge slowly, so that it gains no kinetic energy. Then the agent's work goes entirely into potential energy: \(W_\text{ext} = U_f - U_i\). The electric force does the opposite work at every step, so \(W_\text{field} = -(U_f - U_i)\). The two works always add to zero for a slow move.
Take the pair of the example and pull it from \(30\,\text{cm}\) to \(60\,\text{cm}\). The energy halves from \(0.18\,\text{J}\) to \(0.09\,\text{J}\), so \(W_\text{ext} = -0.09\,\text{J}\). The agent does negative work because the repulsion moves the charges apart by itself; the agent only holds them back.
Charges of \(+4\,\mu\text{C}\) and \(-4\,\mu\text{C}\) are \(10\,\text{cm}\) apart. How much work must an outside agent do to separate them slowly to a very large distance?
The starting energy is \(U_i = kq_1q_2/r = 9 \times 10^9 \times 4 \times 10^{-6} \times (-4 \times 10^{-6})/0.1 = -1.44\,\text{J}\), using \(k\).
At a very large distance \(U_f = 0\). So \(W_\text{ext} = U_f - U_i = 0 - (-1.44) = 1.44\,\text{J}\).
The work is positive, as it must be: the agent pulls against the attraction the whole way.
Electric potential energy is stored in the arrangement of charges. It belongs to the whole system, never to one charge alone. When one charge is held fixed, we may speak of the energy "of the other charge" only as a short form.
Keep the two works apart. The outside agent does \(U_f - U_i\); the field does \(U_i - U_f\). Mixing them flips every sign in a problem. Write down whose work you want before you start.
Charges of \(5\,\mu\text{C}\) and \(-2\,\mu\text{C}\) are moved slowly from \(30\,\text{cm}\) apart to \(50\,\text{cm}\) apart. How much work does the outside agent do, in joules?
Energy becomes motion
Now let go. No outside agent acts, and the only force doing work is the conservative Coulomb force. So the mechanical energy of the pair stays constant: \(K + U\) is the same at every separation. As a like pair moves apart, \(U\) falls and \(K\) rises by exactly as much.
In Fig 1.1 the particle starts with \(U = 0.18\,\text{J}\) and \(K = 0\). Very far away \(U = 0\), so all \(0.18\,\text{J}\) has become kinetic energy: \(\tfrac12 mv^2 = 0.18\), which gives \(v^2 = 36\) and \(v = 6\,\text{m/s}\). At \(60\,\text{cm}\) half the energy has changed form, and the particle moves at \(4.2\,\text{m/s}\).
The pith balls of the opening do the same. Pushing them together stored energy in the pair. Released, that energy became the kinetic energy of both balls, shared between them.
The closest approach
Run the motion backwards, as the second chip of Fig 1.1 does. A particle thrown straight at a fixed like charge slows down as \(U\) grows. It stops for an instant where all its kinetic energy has become potential energy. That separation is the distance of closest approach: \(K_\infty = kq_1q_2/r_\text{min}\).
Our \(3\,\mu\text{C}\) particle, thrown in from far away at \(6\,\text{m/s}\), stops exactly \(30\,\text{cm}\) away, where it started before. A \(10\,\text{g}\) particle with \(2\,\mu\text{C}\) thrown at the same speed at the fixed \(2\,\mu\text{C}\) gets closer. It stops where \(kq_1q_2/r = 0.18\,\text{J}\), so \(r = 0.036/0.18 = 0.2\,\text{m}\).
An alpha particle, of charge \(2e\), has a kinetic energy of \(5\,\text{MeV}\). It is fired head-on at a gold nucleus of charge \(79e\), which stays at rest. How close does it get?
\(K = 5 \times 1.6 \times 10^{-13} = 8 \times 10^{-13}\,\text{J}\). Charges \(2e\) and \(79e\), with \(e = 1.6 \times 10^{-19}\,\text{C}\). Route: point charges, so use \(k\).
At the closest point the alpha stops, so \(K = k(2e)(79e)/r_\text{min}\).
\(r_\text{min} = 9 \times 10^9 \times 158 \times (1.6 \times 10^{-19})^2/(8 \times 10^{-13}) = 3.64 \times 10^{-26}/(8 \times 10^{-13}) = 4.5 \times 10^{-14}\,\text{m}\).
The alpha turns back about \(4.5 \times 10^{-14}\,\text{m}\) from the centre of the gold nucleus. The atom is about \(10^{-10}\,\text{m}\) across, so the nucleus must be smaller still.
A \(1\,\text{g}\) particle carrying \(1\,\mu\text{C}\) is released from rest \(10\,\text{cm}\) from a fixed charge of \(4\,\mu\text{C}\). What is its speed when it is very far away, in m/s?
A proton is fired head-on at a fixed nucleus and gets to within a distance \(d\). The proton's kinetic energy is doubled. What is the new distance of closest approach?
Two small spheres each carrying \(1\,\text{C}\) are held \(1\,\text{m}\) apart. Estimate the potential energy of the pair, in joules.
In 1909 Hans Geiger and Ernest Marsden fired alpha particles at a thin gold foil in Rutherford's laboratory in Manchester. A few bounced almost straight back. In 1911 Rutherford explained it with the energy account of this lesson. An alpha particle turns back only if it meets a positive charge packed into a tiny space, where \(kq_1q_2/r\) can match its kinetic energy. The closest approach gave an upper limit to the size of the nucleus.
- Writing \(U = kq_1q_2/r^2\).The energy falls as \(1/r\); the force falls as \(1/r^2\).
- Dropping the signs of the charges.Put each charge in with its sign; the sign of \(U\) carries the physics.
- Confusing the agent's work with the field's work.\(W_\text{ext} = U_f - U_i\) for a slow move; the field does the negative of it.
- Giving the potential energy to one charge.The energy belongs to the pair; "the energy of the charge" is only a short form when the other is fixed.