Fundamentals of Electromagnetics with Engineering Applications
Fundamentals of Electromagnetics with Engineering Applications
1st Edition
ISBN: 9780470105757
Author: Stuart M. Wentworth
Publisher: Wiley, John & Sons, Incorporated
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Chapter 4, Problem 4.1P

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

Expert Solution & Answer
Check Mark
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 days .

Explanation of Solution

Given:

The permittivity of polystyrene is 2.56 and its conductivity is 1017 .

Concept used:

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

  ρv=ρ0eσεt  ......(1)

Here, ρv is the volume charge density, ρ0 is initial charge density at t=0 .

Calculation:

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

  ρv=0.01ρ0

Substitute 0.01ρ0 for ρv , 2.56 for ε and 1017 for σ in equation (1).

  0.01ρ0=ρ0eσ ε r ε 0 t0.01=e 10 17 2.56( 8.854× 10 12 )tln(0.01)= 10 172.56( 8.854× 10 12 )tt=2.56(8.854× 105)ln(0.01)

Simplify the above equation.

  t=22.66(4.605)×105=104.37×105 s

Convert the time in seconds to days.

  t=104.37×105 s1 day86400 s=0.0012007×105 days120 days

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 days .

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Fundamentals of Electromagnetics with Engineering Applications

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