Question

In places such as hospital operating rooms or factories for electronic circuit boards, electric sparks must be avoided. A person standing on a grounded floor and touching nothing else can typically have a body capacitance of 150 pF , in parallel with a foot capacitance of 80.0 pF produced by the dielectric soles of his or her shoes. The person acquires static electric charge from interactions with his or her surroundings. The static charge flows to ground through the equivalent resistance of the two shoe soles in parallel with each other. A pair of rubber-soled street shoes can present an equivalent resistance of 5.00 ×10³ Mω. A pair of shoes with special static-dissipative soles can have an equivalent resistance of 1.00Mω . Consider the person's body and shoes as forming an R C circuit with the ground.(a) How long does it take the rubber-soled shoes to reduce a person's potential from 3.00×10³ V to 100V?

Answer 1

Time taken by the rubber-soled shoes to reduce a person's potential from 3.00×10³ V to 100V is 3.91s

The initial potential difference $\Delta V_o$ is related to the final potential difference $\Delta V$ by

$\Delta V=\Delta V_{o} e^{-t / \tau}$

Our target is to find t, so we rearrange equation (1) for t to be

$\Delta V=\Delta V_{o} e^{-t / RC}\\ \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) = \frac{-t}{RC}\\ t= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \tag{2}$

The body has a capacitance 150 pF while the foot has a capacitance 80 pF and both are in parallel connection. So, the equivalent capacitance is

$C= 150 \mathrm{~pF} + 80 \mathrm{~pF} = 230 \mathrm{~pF}$

The shoes of the rubber-soled has resistance $R=5000 \mathrm{~M\Omega}$. Plug the values for $\Delta V, \Delta V_o$ and C into equation (2) to get the time t.

$t&= -RC \ln \left ( \frac{\Delta V}{\Delta V_{o}} \right ) \\ &= -( 5000 \times 10^{6} \mathrm{~\Omega})(230 \times 10^{-12} \mathrm{~F}) \ln \left ( \frac{100\mathrm{~V}}{3000 \mathrm{~V}} \right ) \\ &= {3.91 \mathrm{~s}}$

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