Question

In figure 25-29, a potential difference V = 100 Vis applied across a capacitor arrangement with capacitances ${\mathit{C}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathbf{F}$ , ${\mathit{C}}_2\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{}\mathit{\mu }\mathbf{F}$ and ${\mathit{C}}_3\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathit{\mu }\mathbf{F}$. (a)What is charge ${\mathit{q}}_{\mathbf{3}}$ ?(b) What is potential difference ${\mathit{V}}_{\mathbf{3}}$ ?(c)What is stored energy${\mathit{U}}_{\mathbf{3}}$for capacitor 3?(d)What is ${\mathit{q}}_{\mathbf{1}}$ ?(e) What is ${\mathit{V}}_{\mathbf{1}}$ ?(f) What is${\mathit{U}}_{\mathbf{1}}$for capacitor 1?(g) What is ${\mathit{q}}_{\mathbf{2}}$ ?(h) What is ${\mathit{V}}_{\mathbf{2}}$ ?(i) What is ${\mathit{U}}_{\mathbf{2}}$ for capacitor 2?

Short answer

1. Charge $q_3$is $7.50x{10}^{-4}\mathrm{C}$
2. Potential difference $V_3$is 50 V .
3. Stored energy $U_3$is $1.88\times {10}^{-2}J$
4. Charge $q_1$is $5.00\times {10}^{-4}C$.
5. Potential difference $V_1$is 50 V
6. Stored energy $U_1$isJ
7. Charge $q_2$is $2.50\times {10}^{-4}C$
8. Potential difference $V_2$is 50 V.
9. Energy stored $U_2=6.25\times {10}^{-3}J$is

Step-by-step solution

The given data

1. Potential difference, V = 100 V
2. Capacitance,$C_1=10.0\mu F$
3. Capacitance,$C_2=5.00\mu F$
4. Capacitance,$C_3=15.00\mu F$

Understanding the concept of the charge and capacitance

We use the relation of charge and capacitance to find the charge on the capacitor. And using the formula of energy stored in the capacitor in the electrostatic field, we can calculate the energy stored for the capacitor.

Formulae:

The energy stored between the capacitor plates, $U=\frac{1}{2}C{V}^2$ ...(i)

The charge stored between the plates of the capacitor, q = CV ...(ii)

The equivalent capacitance of a parallel connection of capacitors,

$C_{equivalent}=\sum _{}\frac{1}{C_i}$ ...(iii)

The equivalent capacitance of a series connection of capacitors,

$\frac{1}{C_{equivalent}}=\sum _{}\frac{1}{C_i}$ ...(iv).

(a) Calculation of the charge q3

The potential across $C_1$is same as across $C_2$.

We first find the equivalent capacitance of the capacitor 1 and 2 first. They are parallel. So, their equivalent capacitance is given using equation (iii) as:

$C_{12}=C_1+C_2$

Now, $C_{12}$and $C_3$are in series. The equivalent capacitance of both is given suing equation (iv) by,

$$\begin{gathered} \frac{1}{C_{123}}=\frac{1}{C_3}+\frac{1}{C_1+C_2} \\ {C}_{123}=\frac{\left(C_1+C_2\right)C_3}{C_1+C_2+C_3} \end{gathered}$$ ...(v)

Now, charge across $C_{123}$is given using equation (ii) as follows:

(For, $V_{123}=V$, $Q_{123}=Q$)

$C_{123}=\frac{Q}{V}$ …(vi)

This charge Q will appear on $C_{12}$and $C_3$, that is given by using equation (ii) as:

$$\begin{gathered} V_1=\frac{Q}{C_{12}} \\ =\frac{Q}{C_1+C_2} \\ =\frac{C_{123}V}{C_1+C_2}\left(bysubstitutingchargefromequation\left(vi\right)\right) \\ =\frac{\left(C_1+C_2\right)C_3}{\left(C_1+C_2+C_3\right)}\frac{V}{C_1+C_2}\left(fromequation\left(v\right)\right) \\ =\frac{C_{3}V}{C_1+C_2+C_3} \\ =\frac{\left(15\mu F\right)\left(100V\right)}{10\mu F+5\mu F+15\mu F} \\ =50V \end{gathered}$$

Now, for theparallel connection of capacitors, the potential difference values of both capacitors are same. Thus,

$$\begin{gathered} V_2=V_1 \\ =50V \end{gathered}$$

And the potential difference across the capacitor 3 is given by,

$$\begin{gathered} V_3=V-V_1 \\ =100V-50V \\ =50V \end{gathered}$$

Now, the charge of capacitor 1 is given using equation (ii) as:

$$\begin{gathered} q_1=(10.0\mu F)\left(50V\right) \\ =5.00\times {10}^{-4}\mathrm{C} \end{gathered}$$

Similarly, the charge of capacitor 2 is given using equation (ii) as:

$$\begin{gathered} q_2=(5.00\mu F)\left(50V\right) \\ =2.50\times {10}^{-4}\mathrm{C} \end{gathered}$$

Similarly, the charge of capacitor 3 is given as:

$$\begin{gathered} q_3=q_1+q_2 \\ =7.50\times {10}^{-4}C \end{gathered}$$

Hence, the value of the charge is $7.50\times {10}^{-4}C$.

(b) Calculation of the potential difference V3

From the calculations of part (a), we get the value of potential difference on the capacitor 3 is 50 V .

(c) Calculation of the stored energy in the capacitor 3, U3

Substituting the given data, we can get the value of the stored energy between the capacitor plates 3 using equation (i) as follows:

$$\begin{gathered} U_3=\frac{1}{2}\left(15.0\times {10}^{-6}C\right){\left(50V\right)}^2 \\ =1.88\times {10}^{-2}J \end{gathered}$$

Hence, the value of the energy is $1.88\times {10}^{-2}J$.

(d) Calculation of the charge q1

From the calculations of part (a), the value of the charge between the capacitor plates 1 is $5.00\times {10}^{-4}C$.

(e) Calculation of the potential difference V1

From the calculations of part (a), we can get the value of potential difference between the capacitor plates 1 is 50 V.

(f) Calculation of the stored energy U1

Substituting the above values in equation (i), we can get the value of energy stored between the capacitor plates as follows:

$$\begin{gathered} U_1=\frac{1}{2}\left(10.0\mu F\right){\left(50V\right)}^2 \\ =1.25\times {10}^{-2}J \end{gathered}$$

Hence, the value of the stored energy is $1.25\times {10}^{-2}J$.

(g) Calculation of the charge q2

From the calculations of part (a), the value of the charge between the capacitor plates 2 is $2.50\times {10}^{-4}C$.

(h) Calculation of the potential difference V2

From the calculations of part (a), we can get the value of potential difference between the capacitor plates 2 is 50 V.

(i) Calculation of the stored energy, U2

Substituting the given data in equation (i), we can get the value of the stored energy between the capacitor plates 2 as follows:

$$\begin{gathered} U_1=\frac{1}{2}\left(5.00\mu F\right){\left(50V\right)}^2 \\ =6.25\times {10}^{-3}J \end{gathered}$$

Hence, the value of the stored energy is $6.25\times {10}^{-3}\mathrm{J}$.