Problem 90 - College Physics

Question

A 4600\(\Omega\) resistor is connected across a charged 0.800 \(\mathrm{nF}\) capacitor. The initial current through the resistor, just after the connection is made, is measured to be 0.250 A. (a) What magnitude of charge was initially on each plate of this capacitor? (b) How long after the connection is made will it take before the charge is reduced to 1\(/ e\) of its maximum value?

Step-by-Step Solution

Answer

Initial charge is \(9.2 \times 10^{-7} \text{ C}\). Time to reduce charge is \(3.68 \times 10^{-6} \text{ s}\).

Step 1: Initial Charge Calculation

The current through the resistor just after the connection is made can be expressed using Ohm's Law: \( I = \frac{V}{R} \), where \( I = 0.250 \text{ A} \) and \( R = 4600 \Omega \). Rearranging for voltage, \( V = I \times R = 0.250 \times 4600 = 1150 \text{ V} \). The charge on the capacitor can be deduced using the formula \( Q = C \times V \), where \( C = 0.800 \times 10^{-9} \text{ F} \). Thus, \( Q = 0.800 \times 10^{-9} \times 1150 = 9.2 \times 10^{-7} \text{ C} \).

Step 2: Time Constant Calculation

The time constant \( \tau \) for an RC circuit is calculated using \( \tau = R \times C \). Here, \( R = 4600 \Omega \) and \( C = 0.800 \times 10^{-9} \text{ F} \), so \( \tau = 4600 \times 0.800 \times 10^{-9} = 3.68 \times 10^{-6} \text{ s} \).

Step 3: Time to Reduce Charge to 1/e

The charge on the capacitor decays over time according to \( Q(t) = Q_0 e^{-t/\tau} \). To find when the charge reduces to \( \frac{1}{e} \) of its initial value \( Q_0 \), set \( Q(t) = \frac{Q_0}{e} \). Solving \( e^{-t/\tau} = \frac{1}{e} \) gives \( t = \tau \). Therefore, \( t = 3.68 \times 10^{-6} \text{ s} \).

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in the study of electrical circuits. It states that the current \( I \) through a conductor between two points is directly proportional to the voltage \( V \) across the two points, with the resistance \( R \) being the constant of proportionality. This relationship is expressed as:

• \( I = \frac{V}{R} \)

To find the initial voltage across the resistor in our exercise, we rearrange the formula to solve for \( V \):

• \( V = I \times R \)

Given the current \( I = 0.250 \text{ A} \) and resistance \( R = 4600 \Omega \), we calculated the voltage as \( 1150 \text{ V} \). This demonstrates how we can apply Ohm's Law to determine unknown values in an electrical circuit. The simplicity of Ohm's Law makes it incredibly practical for analyzing basic RC circuits, like the one in our problem.

Capacitor Charge

Capacitors are essential components in electrical circuits, capable of storing and releasing electrical energy. The charge \( Q \) on a capacitor is related to the voltage \( V \) across it and its capacitance \( C \) through the relationship:

• \( Q = C \times V \)

Capacitance \( C \) is a measure of a capacitor's ability to store charge per unit voltage and is expressed in farads (F). In this problem, we have a 0.800 nF (nanofarad) capacitor.
Using the initial voltage of 1150 V determined from Ohm's Law, we calculate the initial charge \( Q \) as:

• \( Q = 0.800 \times 10^{-9} \times 1150 = 9.2 \times 10^{-7} \text{ C} \)

This formula reflects how the charge stored in a capacitor is proportional to both its capacitance and the voltage applied across it.

Time Constant

The time constant \( \tau \) is a crucial concept in the analysis of RC circuits. It determines how quickly a capacitor charges or discharges through a resistor. The time constant is calculated as:

• \( \tau = R \times C \)

where \( R \) is the resistance and \( C \) is the capacitance. In our example, we calculate:

• \( \tau = 4600 \times 0.800 \times 10^{-9} = 3.68 \times 10^{-6} \text{ s} \)

This tells us that in approximately 3.68 microseconds, the charge on the capacitor will decrease to about 37% of its initial value (or reach \( 1/e \) of its initial value). Understanding the time constant helps in predicting how fast or slow a capacitor reacts in a circuit.

Exponential Decay

Exponential decay describes the process where the charge on a capacitor decreases at a rate proportional to its current value. In an RC circuit, the charge \( Q(t) \) at time \( t \) can be described by:

• \( Q(t) = Q_0 e^{-t/\tau} \)

where \( Q_0 \) is the initial charge on the capacitor, \( e \) is the base of natural logarithms, and \( \tau \) is the time constant. This equation shows how the charge decreases exponentially over time.
For our specific task, we are interested in when the charge reduces to \( \frac{1}{e} \) of its initial value, which occurs at \( t = \tau \). Thus, using the time constant calculated as \( 3.68 \times 10^{-6} \text{ s} \), we find that this is the time at which the charge is about 37% of its maximum value. Exponential decay illustrates the non-linear, smoothly decreasing nature of capacitor discharge over time.