5. Electrostatic effects
Discharging a capacitor P.25
 We have seen that a capacitor loses its charge exponentially. In mathematical terms, this means that the charge gets closer and closer to zero without reaching it. So we can’t measure how long it takes to fully discharge. However, we can measure how long it takes to reach, for the moment, half of the starting charge.
 Choosing a time interval We have been calculating tables of figures using a step-by-step method. We have to choose the time interval carefully. If we make it too small, then it will take lots of steps to get a decent graph. However, if we make it too big, then the whole system becomes unrealistic. For example, if we were to choose a time interval of 1 second, then this would tell us that 250 mC would leave the capacitor in the first second. However, the capacitor only has 20mC on it to start with. So a step size of 1 second gives a ridiculous and unrealistic result.
Measuring discharge
On page 24, we saw what happens when a 2 mF capacitor discharges through a 40 W resistor. It took about 50 ms to reach half of its starting charge. The table below has the same calculation that we used on page 24. However, this time, you can change the values of resistance and capacitance – you may also have to change the value of the time interval (see box).

Try different values of R and C. What is the half-life in each case? Is there a pattern between half-life and resistance and half-life and capacitance. (Try doubling, tripling, halving etc R or C whilst keeping the other one fixed; this will help you find patterns.)

 Capacitance mF choose a value between 0.1 and 100 mF Resistance W choose a value between 1 and 1000 W Time interval s
 time /s charge /mC voltage /V current /A charge flow /mC
Picture 5.15 Table of discharge values highlighting what's left after RC

Results
You should have found that:
• half-life µ R
• half-life µ C

We can say that Half-life µ R.C

 Making sense of the effects of R and C Increasing resistance If we double the resistance, then the current will always be half as much. This means that it will take twice as long for the same amount of charge to flow. Therefore, it will take twice as long for half the charge to decay away. So doubling the resistance doubles T1/2. Increasing the capacitanceIf we double the capacitance and have the same starting voltage, then there will be twice as much charge to start off with. It will take twice as long for half of it to flow off the plates. So doubling the capacitance doubles T1/2.
 Picture 5.14 The charge drops to a half in 69% of RC. It drops to 37% in a time of RC.
 Time constant The half life is proportional to the product RC. In fact, the half life equals 69% of RC. We call the product RC the time constant for the circuit. It has the symbol t: t = RC And rather than measuring the half life (69% of the time constant), we measure how much the capacitor discharges in a time equal to the time constant. This is 37% of the starting value. I.e. instead of measuring the time to fall to a half (as in radioactivity), we measure the time it takes to fall to 37%. The charge on a capacitor falls to 37% of its starting value for every time constant (equal to RC). What is special about 37%? It corresponds to 1/e where e is a natural number like p. The natural number e = 2.718 (to 4 significant figures).

 Question 20 We measure the time it takes for a capacitor to discharge through a resistor using the time constant. a) Show that the ohm is equivalent to the units of J C-2 s Click shift/return to get a line break in your answer b) Show that the farad is equivalent to the units C2 J-1 c) What are the equivalent units of the time constant, RC. d) How much has a capacitor discharged through a resistor after i. one time constant, RC. ii. two time constants, 2.RC?

 Summary                                           Close the discharge of a capacitor through a resistor is measured by the time constant, t t = RC in one time constant, the charge falls to 37% of its starting value