Question

75. A $300\mu \mathrm{F}$capacitor is charged to $9.0\mathrm{V}$, then connected in parallel with a $5000\Omega$resistor. The capacitor will discharge because the resistor provides a conducting pathway between the capacitor plates, but much more slowly than if the plates were connected by a wire. Let $t=0\mathrm{s}$ be the instant the fully charged capacitor is first connected to the resistor. At what time has the capacitor voltage decreased by half, to $4.5\mathrm{V}$?
Hint: The current through the resistor is related to the rate at which charge is leaving the capacitor. Consequently, you'll need a minus sign that you might not have expected.

Short answer

The potential difference will be decreased by half at $t=1.0\mathrm{s}$.

Step-by-step solution

Given information

A $300\mu \mathrm{F}$capacitor is charged to $9.0\mathrm{V}$, in parallel with a $5000\Omega$ resistor. The capacitor will discharge because the resistor provides a conducting pathway between the capacitor plates, but much more slowly than if the plates were connected by a wire.

Explanation

At $t=0$the capacitor is associated with the resistor, and at that point, the potential difference at the capacitor is $\mathrm{\Delta }V\left(0\right)=\mathrm{\Delta }{V}_0=9.0\mathrm{V}$. So the capacitor functions as a battery for the resistor. Still, the current is streaming via the resistor, the charge gets moved from the positive plate of the capacitor to the negative plate, which reduces the charge of the plates and, therefore, the potential difference on the capacitor. In other words $\mathrm{\Delta }V=\mathrm{\Delta }V\left(t\right)$ it is a function of time.

Recall that the current is similar to the rate of flow of the charge, but since the current takes out the positive charge from the positive plate, take the minus sign in front.

$I=-\frac{dq}{dt}$

From Ohm's law, $I=\frac{\mathrm{\Delta }V}{R}$

$-\frac{dq}{dt}=\frac{\mathrm{\Delta }V}{R}$(a)

Explanation

The capacitor is $\mathrm{\Delta }V=\frac{q}{C}$,

$d\left(\mathrm{\Delta }V\right)=\frac{1}{C}dq\Rightarrow dq=Cd\left(\mathrm{\Delta }V\right)$

From the equation (a),

$-\frac{1}{C}\frac{d\left(\mathrm{\Delta }V\right)}{dt}=\frac{\mathrm{\Delta }V}{R}$

$\Rightarrow \frac{d\left(\mathrm{\Delta }V\right)}{\mathrm{\Delta }V}=-\frac{1}{RC}dt$

Integrating this equation, using the fact that $t=0$, then $\Delta V=\Delta V_0$and, $t=\tau$then,$\Delta V=\frac{\Delta V_0}{2}$ ,
${\int }_{\Delta V_0}^{\frac{\Delta V_0}{2}}\frac{d\left(\Delta V\right)}{\Delta V}=-\frac{1}{RC}{\int }_0^{\tau }dt$

$t\Rightarrow \ln \frac{\frac{\Delta V_0}{2}}{\Delta V_0}=-\frac{1}{RC}\tau$

$\ln \frac{1}{2}=-\frac{1}{RC}\tau$

$\tau =RC\ln 2=1.0\mathrm{s}$

The potential difference will be decreased by half at$t=1.0s$