Chapter 4, Problem 4.1P

How long does it take for charge density to drop to 1% of its initial value in polystyrene?

To determine

The time taken by polystyrene for its charge density to drop to $1\%$ of its initial value.

Answer to Problem 4.1P

The time taken by polystyrene for its charge density to drop to $1\%$ of its initial value is $120\text{ days}$ .

Explanation of Solution

Given:

The permittivity of polystyrene is $2.56$ and its conductivity is ${10}^{-17}$ .

Concept used:

The point form of current continuity equation can be written as,

  ${\text{ρ}}_{\text{v}}={\text{ρ}}_0{e}^{-\frac{\sigma }{\epsilon }t}$  ......(1)

Here, ${\text{ρ}}_{\text{v}}$ is the volume charge density, ${\text{ρ}}_0$ is initial charge density at $t=0$ .

Calculation:

The $1\%$ of the initial charge density of polystyrene is,

  ${\text{ρ}}_{\text{v}}=0.01{\text{ρ}}_0$

Substitute $0.01{\text{ρ}}_0$ for ${\text{ρ}}_{\text{v}}$ , $2.56$ for $\epsilon$ and ${10}^{-17}$ for $\sigma$ in equation (1).

  $\begin{array}{c}0.01{\text{ρ}}_0={\text{ρ}}_0{e}^{-\frac{\sigma }{{\epsilon }_{\text{r}}{\epsilon }_0}t}\\ 0.01=e^{-\frac{{10}^{-17}}{2.56\left(8.854\times {10}^{-12}\right)}t}\\ \ln \left(0.01\right)=-\frac{{10}^{-17}}{2.56\left(8.854\times {10}^{-12}\right)}t\\ t=-2.56\left(8.854\times {10}^5\right)\ln \left(0.01\right)\end{array}$

Simplify the above equation.

  $\begin{array}{c}t=-22.66\left(-4.605\right)\times {10}^5\\ =104.37\times {10}^5\text{ s}\end{array}$

Convert the time in seconds to days.

  $\begin{array}{c}t=104.37\times {10}^5\text{ s}\cdot \frac{1\text{ day}}{86400\text{ s}}\\ =0.0012007\times {10}^5\text{ days}\\ \approx 120\text{ days}\end{array}$

Therefore, the time required is $120$ days.

Conclusion:

Thus, the time taken by polystyrene for its charge density to drop to $1\%$ of its initial valueis $120\text{ days}$ .