Question

The charge flowing through a conductor varies with time as $Q = at - b{t^2}$ . Then, the current:
(A) Decreases linearly with time.
(B) Reaches a maximum and then decreases.
(C) Falls to zero at time $t = \dfrac{a}{{2b}}$
(D) Changes at a rate $- 2b$

Answer 1

Hint : Here, the charge flowing through a conductor varies with time such that the current is given by the rate of change in charge flowing through the conductor. Thus, we will see the changes in the current that if it increases or decreases or remains constant.

Complete Step By Step Answer:
According to given data in the above question is that the charge flowing through the conductor varies with time as:
$Q = at - b{t^2}$
Current is the rate of change of charge in the conductor, thus, we have
$I = \dfrac{{dQ}}{{dt}}$
$\Rightarrow I = \dfrac{{d\left( {at - b{t^2}} \right)}}{{dt}} = a - 2bt$
That means the current is linearly decreasing with time.
For $I = 0$
$\Rightarrow a - 2bt = 0$ or $t = \dfrac{a}{{2b}}$
Thus, current rate is $\dfrac{{dI}}{{dt}} = - 2b$
Thus, the rate of change in current will be constant.
Therefore, the correct answer is option C and D both.

Note :
We know that the current is the rate of change of charge and we have used this concept to obtain the answer. The current is changing linearly but also decreasing with time. And also we have found that the current changes or decreases with the rate of $- 2b$ . Take care of the differentiation we tend to use in this problem.