91Ó°ÊÓ

Rate of charge flow $=dQ/dt$

Capacitor$=10\mathrm{μ}F$

Voltagerole="math" localid="1648643337782" $=2.0V/S$

We know from the definition that capacitance is the ratio of the potential charge to the difference between the potentials across the capacitor. That is,

$C=\frac{q}{U}$

If we divide both the denominator and numerator by the time, we can write

$C=\frac{q/t}{U/t}$

or, more specifically, if we take the infinitesimal changes of the charge and potential difference, we can write

$C=\frac{dq/dt}{dU/dt}$

Let us not forget that the change of charge per unit time is nothings less but the current ; that is,

$I:=\frac{dq}{dt}.$

This means that we can write the current as

$I=C\frac{dU}{dt}$

We know the capacitance, and we know the rate of change of the potential difference as well.

Then the numerical solution will be

localid="1648643685911" $I=(10\mathrm{μ}F·{10}^{-6}×2.0V/S)$

Multiply the expression,

$=20\mathrm{μ}\mathrm{A}$

Hence, the current to a $10\mathrm{μ}F$capacitor whose voltage is increasing at the rate of $2.0\mathrm{V}/\mathrm{S}$is $20\mathrm{μ}\mathrm{A}$.