**How many coulombs of charge pass a cross-section of the wire in the time interval between $t=0\,s$ and $t=8.5\,s$?****Express your answer using two significant numbers.****What constant current would transport the same charge in the same time interval?****Express your answer using two significant numbers.**

The primary objective of this problem is to calculate the amount of charge that could pass through a cross-section in the given interval of time, as well as the constant current that will transfer the charge.

Electric charge is a vital property of matter carried by certain fundamental particles which govern how the particles react to a magnetic or electric field. Electric charge can be either negative or positive and appears in precisely defined natural units and cannot be created or destroyed. It is therefore conserved.

**Expert Answer**

To begin with this problem, use integration to determine the charge that passes through the cross-section during the given time interval. Then, using the relationship between current, time interval, and charge, calculate the current.

The given equation of current can be plotted against time as:

1- Given

Electric current $I=55A-\left(0.65\dfrac{A}{s^2}\right)t^2$

Initial time $t_1=0\,s$

Final time $t_2=8.5\,s$

The charge that passes through a cross-section in a given time interval is

$Q=\int\limits_{t_1}^{t_2}\,I dt$

$Q=\int\limits_{0\,s}^{8.5\,s}\,\left(55A-\left(0.65\dfrac{A}{s^2}\right)t^2\right) dt$

$Q=[55t\,A]_{0\,s}^{8.5\,s}-\left[\dfrac{0.65}{3}\dfrac{A}{s^2}\cdot t^3\right]_{0\,s}^{8.5\,s}$

$Q=467.5\,C-133.06\,C$

$Q=334.44\,C$

( where $C=As$ )

Consequently, the amount of charge that passes through a cross-section in the given time interval is $334.44\,C$.

2- The following equation gives the constant current.

$I=\dfrac{\Delta Q}{\Delta t}$

Because the amount of charge is the same in the given interval, therefore, $\Delta Q=Q$ and

$I=\dfrac{Q}{t_2-t_1}$

In the above equation, substitute the given values for $Q$, $t_1$, and $t_2$.

$I=\dfrac{334.44\,C}{8.5\,s-0\,s}$

$=39.35\,A$

( where $A=\dfrac{C}{s}$ )

Hence, the constant current required to transport the charge is $39.35\, A$.

Consider an example to obtain a charge amount using the separation of variables method.

**Example 1**

What will be the amount of charge (in Coulombs) through the cross-section of a wire in the interval $t_1=2\,s$ to $t_2=6\,s$ when the current is expressed by the equation $I=3t^2-2t+1$?

Given

$I=3t^2−2t+1$

Since

$I=\dfrac{dQ}{dt}$

(Because $\Delta$ represents the finite variability of a quantity, therefore, we have replaced $\Delta $ by $d$.)

$dQ=I\,dt$

$\int dQ=\int\limits_{2}^{6}(3t^2−2t+1)\,dt$

$Q=\left[\dfrac{3t^3}{3}-\dfrac{2t^2}{2}+t\right]_2^6$

$Q=\left[ (216-8)- (36-4)+(6-2)\right] $

$Q=180\,C$

**Example 2**

A car battery generates $530\, C$ of charge in $6\, s$ when its engine is started, what will be the current $(I)$?

Since,

$I = \dfrac{\Delta Q}{\Delta t}$

Substituting the values for time and charge in the above formula of current yields

$ I = \dfrac{\Delta Q}{\Delta t}=\dfrac{530\,C}{6\,s}=88.33\,\dfrac{C}{s} $

$I=88.33\,A$

* Images/mathematical drawings are created with GeoGebra.*