Chapter 19, Problem 46P

(a)

To determine

Whether the capacitors are connected in series or parallel.

Answer to Problem 46P

The two capacitors are connected in the parallel connection.

Explanation of Solution

Introduction:

The combination of the capacitors consider as the parallel combination if the potential difference across all the capacitors is same. If the charge through the each capacitor remains same then combination is called the series combination of capacitors.

One end of all the capacitors is connected by the positive terminal of the battery and other end of two capacitors connected to negative terminal of the battery, so the potential difference across two plates remain same. As the potential difference across the two capacitors plate is same, so both capacitors connected in the parallel connection.

Conclusion:

Thus, the two capacitors are connected in the parallel connection.

(b)

To determine

The equivalent capacitance of the capacitors.

Answer to Problem 46P

The equivalent capacitance as a function area of plates and distance between the plates is

  $\epsilon A\left(\frac{1}{d_1}+\frac{1}{d_2}\right)$ .

Explanation of Solution

Given info:

The distance between the middle plate and upper plate is, $d_1$ .

The distance between the middle plate and lower plate is, $d_2$ .

The area of each plate is, $A$ .

Formula Used:

The expression to calculate the capacitance of first capacitor is given as,

  $C_1=\frac{\epsilon A}{d_1}$

Here,

  $\epsilon$ is the permittivity of space between the plates.

The expression to calculate the capacitance of second capacitor is given as,

  $C_2=\frac{\epsilon A}{d_2}$

The expression to calculate the charge of first capacitor is,

  $Q_1=C_{1}V$

Here,

  $V$ is the potential difference across each plates.

The expression to calculate the charge of second capacitor is,

  $Q_2=C_{2}V$

The expression to calculate the total charge of the equivalent connection is,

  $Q=CV$

The capacitors are connected in parallel, so the potential across each capacitor is same and total charge of both capacitors is equal to sum of charge of each capacitor.

The expression to calculate the total charge of the both capacitors in parallel is given by,

  $Q=Q_1+Q_2$     ..........(1)

Calculation:

Substitute the given values in the expression (1).

  $\begin{array}{c}CV=C_{1}V+C_{2}V\\ C=C_1+C_2\\ C=\frac{\epsilon A}{d_1}+\frac{\epsilon A}{d_2}\\ C=\epsilon A\left(\frac{1}{d_1}+\frac{1}{d_2}\right)\end{array}$

Conclusion:

Thus, the equivalent capacitance as a function area of plates and distance between the plates is $\epsilon A\left(\frac{1}{d_1}+\frac{1}{d_2}\right)$ .