Chapter 19, Problem 48P

To determine

The value of the unknown capacitance.

Answer to Problem 48P

The value of the unknown capacitor is $5.14\text{pF}$ .

Explanation of Solution

Given info:

The capacitance of known capacitor is, $C_1=185\text{pF}$ .

The emf of the battery is, $V=25\text{V}$ .

The charge of the known capacitor is, $Q=125\text{pC}$ .

Formula Used:

The expression to calculate the potential of known capacitor is,

  $V_1=\frac{Q}{C_1}$

Here,

• $Q$is the charge and the value of charge of capacitors in series combination remains same.

The expression to calculate the potential of unknown capacitor is,

  $V_2=\frac{Q}{C_2}$

The expression to calculate the emf of the capacitor in series combination is,

  $\begin{array}{l}V=V_1+V_2\\ V=\frac{Q}{C_1}+\frac{Q}{C_2}\end{array}$  ..........(1)

Calculation:

Substitute all the values in the expression (1).

  $\begin{array}{c}25\text{V}=\frac{125\text{pC}}{185\text{pF}}+\frac{125\text{pC}}{C_2}\\ \frac{125\text{pC}}{C_2}=25\text{V}-\frac{125\text{pC}}{185\text{pF}}\\ C_2=\frac{125\text{pC}}{\left(25\text{V}-\frac{125\text{pC}}{185\text{pF}}\right)}\\ =5.14\text{pF}\end{array}$

Conclusion:

Thus, the value of the unknown capacitor is $5.14\text{pF}$ .