91Ó°ÊÓ

The current in a wire at time is given by the expression $I=(2.0\mathrm{A})e^{-A[2.0\mathrm{μ}\mathrm{s})}$, where $t$indicates microseconds and $I$indicates the current intensity.

For a time t that starts at zero and increases, this question shows how current,$I$, is related to time, $t$.

$I=\left(2.0\mathrm{A}\right)e^{\frac{−t}{2.0\mathrm{μ}\mathrm{s}}}$

Find the charge, $Q$, as a function of time where at $t=0\mathrm{μ}\mathrm{s},Q=0C$

A charge is the integral of current with respect to time,

$Q=∫Idt$

Substitute $(2.0\mathrm{A})e^{\frac{-t}{2.0\mathrm{μ}\mathrm{s}}}$for $Q,$

$Q=∫(2.0\mathrm{A})e^{\frac{-t}{2.0\mathrm{μ}\mathrm{s}}}dt$

$=-(2.0\mathrm{A})(2.0\mathrm{μ}\mathrm{s})e^{\frac{-t}{2.0\mathrm{μ}\mathrm{s}}}$

$=(-4.0\mathrm{μ}\mathrm{C})e^{\frac{-t}{2.0\mathrm{μ}\mathrm{s}}}+C$

Here, $C$is the constant of integration.

Solve for $C$by plugging in the initial conditions of charge and time,

$0=(-4.0\mathrm{μ}\mathrm{C})e^{\frac{-0}{2.0\mathrm{μ}\mathrm{s}}}+C$

$=(-4.0\mathrm{μ}\mathrm{C})\left(1\right)+C$

Solve for $C$to get,

$C=4.0\mathrm{A}×\mathrm{μ}\mathrm{s}$

Substitute the value of C in above expression,

$Q=(-4.0\mathrm{μ}\mathrm{C})e^{\frac{-t}{2.0\mathrm{μ}\mathrm{s}}}+4.0\mathrm{μ}\mathrm{C}$

Total amount of charge entered the wire at time$t$is$(-4.0\mathrm{μ}\mathrm{C})e^{\frac{-t}{2.0\mathrm{μ}\mathrm{s}}}+4.0\mathrm{μ}\mathrm{C}$

Graph $Q$versus $t$for the interval $0≤t≤10\mathrm{μ}\mathrm{s}$. Where $Q$indicates the charge quantity.

Conspiring this process with a graph will show the ensuing linear relation.

The expression for the charge in the wire a time $t$is

$Q=(-4.0\mathrm{μ}\mathrm{C})e^{\frac{-t}{2.0\mathrm{μ}\mathrm{s}}}+4.0\mathrm{μ}\mathrm{C}$

This problem for the graph of the charge versus time, from $0\mathrm{s}≤t≤10\mathrm{μ}\mathrm{s}$

Graph $Q$versus $t$,for the interval $0≤t≤10\mathrm{μ}S$is,