Chapter 17, Problem 41P

To determine

The value of charges on each capacitor and value of common potential difference will be calculated.

Answer to Problem 41P

The answers are $712V,1.78\times {10}^{-3}C$ and $4.84\times {10}^{-3}C$ .

Explanation of Solution

Given data:

Capacitor $C_1=2.50\times {10}^{-6}F$

Capacitor $C_2=6.80\times {10}^{-6}F$

Final voltage $V_1{}_{{}_{initial}}=857V$

Final voltage $V_2{}_{initial}=652V$

Formula used:

It is known that: $Q=CV$

Where,

  $Q$ is the charge.

  $\text{C}$ is the capacitance of capacitor.

  $V$ is the potential.

Calculation:

Finding the final potential on both capacitor,

  $V_{final}=\frac{C_1{V}_1{}_{{}_{initial}}+C_2{V}_2{}_{{}_{initial}}}{C_1+C_2}$

  $V_{final}=\frac{\left(2.50\times {10}^{-6}F\right)\left(875V\right)+\left(6.80\times {10}^{-6}F\right)\left(652V\right)}{\left(9.30\times {10}^{-6}F\right)}$

  $V_{final}=72V=V_1=V_2$

Calculating the charge on each capacitor,

  $Q_1{}_{{}_{final}}=C_1{V}_{final}$

  $Q_1{}_{{}_{final}}=\left(2.50\times {10}^{-6}F\right)\left(711.95\right)$

  $\therefore Q_1{}_{{}_{final}}=1.78\times {10}^{-3}C$

  $Q_2{}_{{}_{final}}=C_2{V}_{final}$

  $Q_2{}_{{}_{final}}=\left(6.80\times {10}^{-6}F\right)\left(711.95\right)$

  $\therefore Q_2{}_{{}_{final}}=4.84\times {10}^{-3}C$

Conclusion: The potential, and charges on each capacitor are $712V,1.78\times {10}^{-3}C$ and $4.84\times {10}^{-3}C$ , respectively.