Comprehensive Study Guide for Physics Example 40: Parallel Capacitors

Overview of Example 40: Parallel Capacitor Configuration

Example 40 concerns the analysis of a specific electrical circuit consisting of four capacitors connected in a parallel arrangement. The physical properties defined for these capacitors are as follows: the first capacitor is denoted as $C_1 = 4\,\mu F$, the second as $C_2 = 12\,\mu F$, the third as $C_3 = 8\,\mu F$, and the fourth as $C_4 = 6\,\mu F$. It is essential to note that in a parallel connection, the potential difference (voltage) across each individual capacitor is identical to the total potential difference of the group, which can be expressed mathematically as $V_{total} = V_1 = V_2 = V_3 = V_4$.

In addition to the capacitance values, the problem provides a specific energy metric for the third capacitor. The energy stored in the electric field between the plates of the third capacitor ($C_3$) is given as $(256 \times 10\,J)$. This value serves as the primary gateway for calculating the voltage across the entire system, as it links capacitance and energy through the established physical formula for potential energy.

Calculation of Equivalent Capacitance for the Group

The first requirement of the problem is to determine the equivalent capacitance ($C_{eq}$) of the entire set of four capacitors. In a parallel circuit, the equivalent capacitance is found by simply summing the individual capacitances of all capacitors included in the group. This is because the total surface area for charge storage increases when components are added in parallel.

The formula for calculating this value is:

$C_{eq} = C_1 + C_2 + C_3 + C_4$

By substituting the provided numerical values into the equation, we obtain:

$C_{eq} = 4\,\mu F + 12\,\mu F + 8\,\mu F + 6\,\mu F = 30\,\mu F$

This result represents the total storage capacity of the network as if it were a single capacitor.

Determination of Individual and Total Potential Difference

The second objective is to calculate the potential difference for each capacitor as well as the total potential difference of the system. To find these values, we utilize the energy information provided for the third capacitor ($C_3$). The formula relating stored energy ($PE$), capacitance ($C$), and potential difference ($V$) is:

$PE = \frac{1}{2} C V^2$

Rearranging this formula to solve for the potential difference ($V_3$) for the third capacitor gives:

$V_3 = \sqrt{\frac{2 \times PE_3}{C_3}}$

Applying the verbatim values from the transcript where $PE_3 = (256 \times 10\,J)$ and $C_3 = 8\,\mu F$, or $8 \times 10^{-6}\,F$, allows for the calculation of the voltage. Because the capacitors are connected in parallel, the resulting value for $V_3$ is the same for $V_1$, $V_2$, $V_4$, and the total potential difference $V_{total}$.

Quantifying Specific and Total Electric Charge

The third requirement involves calculating the electric charge stored on each individual capacitor ($Q_1, Q_2, Q_3, Q_4$) and the total charge stored by the entire group ($Q_{total}$). The relationship between charge ($Q$), capacitance ($C$), and voltage ($V$) is defined by the equation:

$Q = C \times V$

Individual charges are calculated as follows:

• For the first capacitor: $Q_1 = C_1 \times V_{total}$
• For the second capacitor: $Q_2 = C_2 \times V_{total}$
• For the third capacitor: $Q_3 = C_3 \times V_{total}$
• For the fourth capacitor: $Q_4 = C_4 \times V_{total}$

The total charge ($Q_{total}$) can be found using two different methods to verify accuracy. The first method is to sum the individual charges: $Q_{total} = Q_1 + Q_2 + Q_3 + Q_4$. The second method is to multiply the total equivalent capacitance by the total potential difference: $Q_{total} = C_{eq} \times V_{total}$.

Electric Field Analysis for the First Capacitor

The final task is to calculate the electric field intensity ($E_1$) existing between the plates of the first capacitor ($C_1$). An additional physical parameter is provided for this step: the distance ($d$) between the plates of the first capacitor is precisely $0.4\,cm$.

To perform this calculation, the distance must be converted into the standard unit of meters:

$d = 0.4\,cm = 0.4 \times 10^{-2}\,m$

The electric field is defined as the potential difference divided by the distance between the plates. The formula used is:

$E_1 = \frac{V_1}{d}$

By dividing the potential difference $V_1$ (found in the previous steps) by the distance $0.4 \times 10^{-2}\,m$, one can find the electric field strength in units of $\text{V/m}$ (Volts per meter).