Chapter 17, Problem 52P

(a)

To determine

The total stored energy before the two capacitors are connected.

Answer to Problem 52P

  $\text{1}{\text{.944×10}}^{\text{-4}}\text{J}$

Explanation of Solution

Given:

The capacitor of $\text{2}\text{.70-μF}$ is charged by a $\text{12}\text{.0}\text{V}$ battery and later disconnected by the battery and connected to $\text{4}\text{.00-μF}$ uncharged capacitor.

Formula used:

The potential energy stored in the capacitor is

  $\text{U=}\frac{\text{1}}{\text{2}}{\text{CV}}^{\text{2}}$

Where,

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

  $\text{V}$ is the voltage across the capacitor plates.

Calculation:

The potential energy stored is

  $\text{U=}\frac{\text{1}}{\text{2}}{\text{C}}_{\text{1}}{\text{V}}_{\text{1}}{}^{\text{2}}$

  $\text{U}\text{=}\frac{\text{1}}{\text{2}}\left(\text{2}{\text{.7×10}}^{\text{-6}}\text{F}\right){\left(\text{12}\text{.0}\text{V}\right)}^{\text{2}}$

  $\text{U}\text{=}\text{1}\text{.94}{\text{×10}}^{\text{-4}}\text{J}$

(b)

To determine

The total stored energy after the two capacitors are connected.

Answer to Problem 52P

  $\text{U=7}{\text{.834×10}}^{\text{-5}}\text{J}$

Explanation of Solution

Given:

The capacitor of $\text{2}\text{.70-μF}$ is charged by a $\text{12}\text{.0}\text{V}$ battery and later disconnected by the battery and connected to $\text{4}\text{.00-μF}$ uncharged capacitor.

Formula used:

The charge on capacitor is

  $\text{Q = CV}$

Where,

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

  $\text{V}$ is the voltage across the capacitor plates.

Moreover, the potential energy is given as

  $\text{U = }\frac{\text{1}}{\text{2}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}$

Calculation:

Assume that the charge stored in the capacitor remains constant and after the connection of the capacitors, the voltage is ${\text{V}}^{\text{'}}$ .The charge on the capacitors be ${\text{Q}}_{\text{1}}$ and ${\text{Q}}_{\text{2}}$ respectively.

The total charge on the capacitor is

  ${\text{Q}}_{\text{total}}{\text{=C}}_{\text{1}}\text{V}$

  ${\text{Q}}_{\text{total}}\text{=}\left(\text{2}{\text{.7×10}}^{\text{-6}}\right)\left(\text{12}\text{.0}\right)$

  ${\text{Q}}_{\text{total}}\text{=3}{\text{.24×10}}^{\text{-5}}\text{C}$

But, ${\text{Q}}_{\text{total}}{\text{=Q}}_{\text{1}}{\text{+Q}}_{\text{2}}$ .

The potential is constant. Therefore,

  $\frac{{\text{Q}}_{\text{1}}}{{\text{C}}_{\text{1}}}\text{=}\frac{{\text{Q}}_{\text{2}}}{{\text{C}}_{\text{2}}}$

So,

  $\frac{{\text{Q}}_1}{{\text{C}}_1}=\frac{{\text{Q}}_{\text{total}}{\text{-Q}}_1}{{\text{C}}_2}$

  ${\text{Q}}_1{\text{=Q}}_{\text{total}}\left(\frac{{\text{C}}_1}{{\text{C}}_{\text{2}}{\text{+C}}_{\text{1}}}\right)$

Hence,

  ${\text{Q}}_1\text{=}\left(\text{3}{\text{.24×10}}^{\text{-5}}\right)\left(\frac{\text{2}{\text{.7×10}}^{\text{-6}}}{\text{2}{\text{.7×10}}^{\text{-6}}{\text{+4×10}}^{\text{-6}}}\right)$

  ${\text{Q}}_1\text{=1}{\text{.31×10}}^{\text{-5}}\text{C}$

Similarly,

  ${\text{Q}}_2\text{=3}{\text{.24×10}}^{\text{-5}}\text{-1}{\text{.31×10}}^{\text{-5}}$

  ${\text{Q}}_2\text{= 1}{\text{.93×10}}^{\text{-5}}\text{C}$

The total potential energy is

  $\text{U=}\frac{{\left({\text{Q}}_{\text{1}}\right)}^{\text{2}}}{{\text{2C}}_{\text{1}}}\text{+}\frac{{\left({\text{Q}}_{\text{2}}\right)}^{\text{2}}}{{\text{2C}}_{\text{2}}}$

  $\text{U=}\frac{{\left(\text{1}{\text{.31×10}}^{\text{-5}}\right)}^2}{\text{2}\left(\text{2}{\text{.7×10}}^{\text{-6}}\right)}\text{+}\frac{{\left(\text{1}{\text{.93×10}}^{\text{-5}}\right)}^2}{\text{2}\left({\text{4×10}}^{\text{-6}}\right)}$

  $\text{U=7}{\text{.834×10}}^{\text{-5}}\text{J}$

(c)

To determine

The change in stored energy.

Answer to Problem 52P

  $\Delta \text{U=1}{\text{.161×10}}^{\text{-4}}\text{J}$ .

Explanation of Solution

Given:

The capacitor of $\text{2}\text{.70-μF}$ is charged by a $\text{12}\text{.0}\text{V}$ battery and later disconnected by the battery and connected to $\text{4}\text{.00-μF}$ uncharged capacitor. The initial energy stored is $\text{U=7}{\text{.834×10}}^{\text{-5}}\text{J}$ and the final energy stored is $\text{1}{\text{.944×10}}^{\text{-4}}\text{J}$ .

Formula used:

It is known that, the potential energy stored in the capacitor is

  $\text{U =}\frac{\text{1}}{\text{2}}{\text{ CV}}^2$

Where,

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

  $\text{V}$ is the voltage across the capacitor plates.

Calculation:

The change in energy stored is

  $\Delta \text{U=7}\text{.834}\times {\text{10}}^{-5}\text{-1}{\text{.944×10}}^{\text{-4}}$

  $\Delta \text{U=1}{\text{.161×10}}^{\text{-4}}\text{J}$