Suppose that, while you are sitting in a chair, charge separation between your clothing and the chair puts you at a potential of 200 V, with the capacitance between you and the chair at 150 pF. When you stand up, the increased separation between your body and the chair decreases the capacitance to 10 pF. (a) What then is the potential of your body? That potential is reduced over time, as the charge on you drains through your body and shoes (you are a capacitor discharging through a resistance). Assume that the resistance along that route is $300G\Omega$. If you touch an electrical component while your potential is greater than 100V, you could ruin the component. (b) How long must you wait until your potential reaches the safe level of 100V?

If you wear a conducting wrist strap that is connected to ground, your potential does not increase as much when you stand up; you also discharge more rapidly because the resistance through the grounding connection is much less than through your body and shoes. (c) Suppose that when you stand up, your potential is 1400 Vand the chair-to-you capacitance is 10pF. What resistance in that wrist-strap grounding connection will allow you to discharge to100V in 0.30 s, which is less time than you would need to reach for, say, your computer?

• a)Potential difference created due to charge separation between the clothing and chair,$V_1=200V$
• b)Capacitance between the chair and the body,$C_1=150pF$
• c)The decreased capacitance value,$C\text{'}=10pF$
• d) Resistance along the route of body to shoes, $R=300G\Omega$
• e) If we touch an electric component with potential difference greater than 100V, the component gets ruined.
• f) Potential difference when the body is standing, $V_d\text{'}=1400V$
• g) The time to discharge , t =0.30s

A condition of potential drop happens due to a change in charge separation between the two ends of the potential bridge because of discharging of the capacitance. This implies that the potential is related to the time constant of a circuit in the same way as the charge is related to it. Thus, using this condition, we calculate the required time and resistance as per the given circumstances.

Formulae:

The charge within the plates of a capacitor,q=CV (i)

The potential difference of a RC circuit,$V=V_0{e}^{-t/RC}$ (ii)

(a)

In the process described in the problem, no charge is gained or lost. Thus, charge value is constant.

Now, using equation (i), the charge between the body and the chair will give the new increased potential on the body as follows:

$$\begin{gathered} C_1{V}_1=C_2{V}_2 \\ {V}_2=\frac{C_1{V}_1}{C_2} \\ =\frac{\left(150pF\right)}{10pF}200V \\ =3000V \end{gathered}$$

Hence, the value of the required potential is 3000V.

(b)

The time constant value of a circuit not only describes the discharging condition of the charge but also the discharging of the potential difference. Thus, the time taken by the body to reach the safe level by discharging from the new potential $V_0=3000V$can be given using equation (ii) as follows:

role="math" localid="1662727335305" $$\begin{gathered} t=RCIn\left(\frac{V_0}{V}\right) \\ =\left(300\times {10}^9\Omega \right)\left(10\times {10}^{-12}F\right)In\left(\frac{3000V}{100V}\right) \\ =10s \end{gathered}$$

This is a longer time than most people are inclined to wait before going on to their next task (such as handling the sensitive electronic equipment).

Hence, the value of time taken is 10s .

(c)

For the condition of standing, the resistance of the wrist-strap grounded connection needed for the potential$V_0=1400V$to drop to the required safe level can be given using equation (ii) as follows:

$$\begin{gathered} R=\frac{t}{CIn\left(V_0/V\right)} \\ =\frac{0.30S}{\left(10\times {10}^{-12}F\right)In\left(1400/100\right)} \\ =1.1\times {10}^{10}\Omega \end{gathered}$$

Hence, the value of the resistance is$1.1\times {10}^{10}\Omega$.