The flash unit of a camera stores its energy in a capacitor. When the unit is switched on, a battery charges the capacitor through a resistor. The charge rises quickly at first, then more and more slowly. The "ready" lamp glows after a few seconds, although the charge never quite reaches its full value. The circuit gives a rule for the rate of charging at each instant. How does a rule for the rate become a formula for the charge itself?
An equation for a rate
Many laws of physics state how fast a quantity changes, not what the quantity is. A radioactive sample loses nuclei at a rate proportional to the number left. A capacitor charges at a rate that falls as its charge grows. An equation that contains a derivative like \(dy/dx\) is called a differential equation. To solve it is to find the function \(y(x)\) itself.
A differential equation is separable when its right side is a product of a function of \(x\) and a function of \(y\): \(dy/dx = f(x)\,g(y)\). To solve it, divide by \(g(y)\) and multiply by \(dx\), so that each side holds one variable. Integrate each side between matching limits: \(y_0\) to \(y\) on one side, and \(x_0\) to \(x\) on the other. Then solve the result for \(y\).
Matching limits do the work of the constant \(C\). The lower limits are the starting values, \(y_0\) at \(x_0\). The upper limits are a later pair, \(y\) at \(x\). So the constant of integration never has to be found separately.
The simplest case is the first equation of motion. The acceleration is a constant, so \(dv/dt = a\). Separate: \(dv = a\,dt\). Integrate from the start, where the velocity is \(u\) at \(t = 0\): \(\int_u^v dv = \int_0^t a\,dt\). This gives \(v - u = at\), so \(v = u + at\). Motion in One Dimension, lesson 6, derives the first equation in exactly these steps.
Put the steps for solving a separable equation in order.
A rate proportional to the amount
In a radioactive sample, each nucleus has the same small chance of decaying in each second. This chance per second is the decay constant \(\lambda\). Among \(N\) nuclei, about \(\lambda N\) decay each second. So the number falls at the rate \(dN/dt = -\lambda N\). The minus sign says that \(N\) decreases. The rate depends on \(N\) itself, so a simple integration of the right side is not possible.
\[ \dfrac{dN}{N} = -\lambda\,dt \]
Why. divide both sides by \(N\) and multiply by \(dt\). Now each side holds one variable.
\[ \int_{N_0}^{N} \dfrac{dN}{N} = -\lambda\int_0^t dt \]
Why. integrate between matching limits: \(N_0\) nuclei at \(t = 0\), and \(N\) nuclei at time \(t\).
\[ \ln\dfrac{N}{N_0} = -\lambda t \]
Why. the integral of \(1/N\) is \(\ln N\), and \(\ln N - \ln N_0 = \ln(N/N_0)\).
\[ N = N_0\,e^{-\lambda t} \]
Read it. take the exponential of both sides, because the exponential undoes the logarithm.
The solution says something that the rate law hides. The fraction left, \(N/N_0 = e^{-\lambda t}\), does not depend on \(N_0\) at all. A large sample and a small sample of the same nuclide lose the same fraction in the same time. The figure shows three samples together.
A sample has a decay constant \(\lambda = 0.3\) per hour. What fraction of its nuclei is left after \(5\) hours? Give a number between 0 and 1.
Approaching a final value
Two more equations of the book have one shape. A steel ball falls through oil. Its weight pulls it down and a drag proportional to its speed holds it back: \(m\,dv/dt = mg - kv\). The drag grows until it balances the weight, at the terminal velocity \(v_t = mg/k\). A capacitor charges through a resistor: \(R\,dq/dt = \mathcal E - q/C\). The charge grows until the current stops, at \(q = C\mathcal E\). Both rates are proportional to the gap left to a final value.
\[ m\dfrac{dv}{dt} = mg - kv = k\,(v_t - v) \]
Why. take \(k\) out as a common factor, with \(v_t = mg/k\), the speed at which the drag balances the weight.
\[ \dfrac{dv}{v_t - v} = \dfrac{dt}{\tau}, \quad \tau = \dfrac{m}{k} \]
Why. separate the variables. The time constant \(\tau\) collects the constants \(m\) and \(k\).
\[ \int_0^{v} \dfrac{dv}{v_t - v} = \int_0^{t} \dfrac{dt}{\tau} \]
Why. the ball starts from rest, so \(v = 0\) at \(t = 0\).
\[ -\ln\dfrac{v_t - v}{v_t} = \dfrac{t}{\tau} \]
Why. substitute \(w = v_t - v\), so \(dw = -dv\); the integral of \(1/w\) gives the logarithm with a minus sign.
\[ v = v_t\left(1 - e^{-t/\tau}\right) \]
Read it. take the exponential of both sides and solve for \(v\).
The quantity \(\tau\) is the time constant. After one time constant the gap has shrunk to \(1/e = 0.37\) of its first size, so the velocity has reached \(63\,\%\) of \(v_t\). The rate is largest at the start, when the gap is largest, and becomes ever smaller as the gap closes. So the final value is approached but never quite reached.
This is a worked example of Circuits: Kirchhoff's Laws, Networks and RC Transients, lesson 5, done with the tool. A \(500\,\mu\text{F}\) capacitor is charged from \(12\,\text{V}\) through \(2\,\text{k}\Omega\). Find the final charge, and the charge and the current at \(t = 1\,\text{s}\). Take \(1/e = 0.37\).
The loop equation is \(R\,dq/dt = \mathcal E - q/C\), which is \(dq/dt = (C\mathcal E - q)/RC\). This is the ball's equation with \(q\) for \(v\), \(C\mathcal E\) for \(v_t\) and \(RC\) for \(\tau\).
Separating as for the ball gives \(q = C\mathcal E\left(1 - e^{-t/RC}\right)\), and \(i = dq/dt = (\mathcal E/R)\,e^{-t/RC}\).
\(RC = 2000 \times 500 \times 10^{-6} = 1\,\text{s}\), so \(t = 1\,\text{s}\) is one time constant. \(C\mathcal E = 6\,\text{mC}\), so \(q = 6 \times 0.63 \approx 3.8\,\text{mC}\). \(\mathcal E/R = 6\,\text{mA}\), so \(i = 6 \times 0.37 \approx 2.2\,\text{mA}\).
The final charge is \(6\,\text{mC}\). At \(1\,\text{s}\) the capacitor holds \(3.8\,\text{mC}\) and \(2.2\,\text{mA}\) flows, as in the physics chapter.
The steel ball in oil has \(v_t = 0.1\,\text{m/s}\) and \(\tau = 12.5\,\text{ms}\). It starts from rest. Find its velocity at \(t = 12.5\,\text{ms}\), in m/s.
A rocket in free space obeys \(m\,dv = -u\,dm\), where \(u\) is the exhaust speed and \(m\) the rocket's mass. Separated and integrated from rest, this gives \(v = u\ln(m_0/m)\). With \(u = 2.5\,\text{km/s}\) and the mass falling to a quarter of \(m_0\), find the final speed in m/s. Use \(\ln 2 = 0.693\).
Which of these equations cannot be solved by separating the variables?
Separation fails in three ways. The right side must be a product \(f(x)\,g(y)\); a sum such as \(x + y\) cannot be separated. Dividing by \(g(y)\) is not allowed where \(g(y) = 0\), and such a value may itself be a solution, like \(N = 0\) for decay. And a logarithm needs a positive argument, so keep the signs: the gap \(C\mathcal E - q\) is positive while the capacitor charges.
The camera flash is now understood. Its loop equation gives the rate of charging, proportional to the charge still missing. Separating and integrating changes this rate into \(q = C\mathcal E(1 - e^{-t/RC})\). The charge reaches \(63\,\%\) of its final value after one time constant and \(99\,\%\) after about five. The ready lamp is set to glow somewhere on that slow final stretch.
Separable equations appear wherever a rate depends on the quantity itself. In Motion in One Dimension, \(dv/dt = a\) gives the first equation of motion. In Radioactivity, \(dN/dt = -\lambda N\) gives exponential decay. In Circuits: Kirchhoff's Laws, Networks and RC Transients, the loop equation gives the charging and discharging of a capacitor. In Viscosity it gives the approach to terminal velocity, and in Momentum and Its Conservation it gives the rocket equation. Newton's law of cooling, the barometric formula and the growth of current in an inductor follow the same steps.
- Integrating the right side of \(dN/dt = -\lambda N\) as if \(N\) were constant.The rate depends on \(N\), so separate first: \(dN/N = -\lambda\,dt\).
- Using mismatched limits, such as \(N_0\) to \(N\) on one side and \(0\) to \(t\) with a different start on the other.The lower limits must be the same instant on both sides, and so must the upper limits.
- Treating the final value as reached after one time constant.After one \(\tau\) the gap is still \(37\,\%\). It takes about five time constants to close to within 1 %.