Question

A potential difference of 300 V is applied to a series connection of two capacitors of capacitances ${\mathit{C}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\mu F}$and ${\mathit{C}}_{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{\mu F}$.What are (a) charge ${\mathbf{q}}_{\mathbf{1}}$ and (b) potential difference ${\mathit{V}}_{\mathbf{1}}$on capacitor 1 and (c) ${\mathit{q}}_{\mathbf{2}}$and (d) ${\mathit{V}}_{\mathbf{2}}$ on capacitor 2? The chargedcapacitors are then disconnected from each other and from the battery. Then the capacitors are reconnected with plates of the same signs wired together (the battery is not used). What now are (e) ${\mathit{q}}_{\mathbf{1}}$ , (f) ${\mathit{V}}_{\mathbf{1}}$ , (g) ${\mathit{q}}_{\mathbf{2}}$ , and (h) ${\mathit{V}}_{\mathbf{2}}$? Suppose, instead,the capacitors charged in part (a) are reconnected with plates of oppositesigns wired together. What now are (i) ${\mathit{q}}_{\mathbf{1}}$, ( j)${\mathit{V}}_{\mathbf{1}}$ , (k)${\mathit{q}}_{\mathbf{2}}$ , and (l)${\mathit{V}}_{\mathbf{2}}$?

Short answer

When potential difference of 300 V is applied to a two capacitors in series then,

a) Charge $q_1$on capacitor $C_1$is $4.80\times {10}^{-4}C$.

b) Potential difference $V_1$across capacitor $C_1$is 240 V.

c) Charge $q_2$on capacitor $C_2$is $4.80\times {10}^{-4}C$.

d) Potential $V_2$on capacitor 2 is 60.0 V .

The charged capacitor are disconnected from each other and from the battery then the capacitors are reconnected with the plates of same signs wired together and battery is not used then,

e) Charge $q_1$on capacitor $C_1$is$1.92\times {10}^{-4}C$.

f) Potential difference $V_1$across capacitor $C_1$is 96.0 V.

g) Charge $q_2$on capacitor $C_2$is $7.68\times {10}^{-4}C$.

h) Potential $V_2$on capacitor 2is 96.0 V

Suppose, instead, the capacitors charged in part (a) are reconnected with plates of opposite signs wired together then,

i) Charge $q_1$on capacitor $C_1$ is 0.00.

j) Potential difference $V_1$across capacitor $C_1$is 0.00.

k) Charge $q_2$on capacitor $C_2$is 0.00.

l) Potential $V_2$on capacitor 2 is 0.00.

Step-by-step solution

The given data

a) Potential difference V=300 V

b) Capacitance $C_1=2.00\mathrm{\mu F}$

c) Capacitance$C_2=8.00\mathrm{\mu F}$

Understanding the concept of the capacitance

The ratio of the quantity of electric charge held on a conductor to an electric potential difference is known as capacitance. We find the equivalent capacitance for series or parallel combinations of capacitors using the equation for series and parallel combinations. Also, we need to use the relation between the charge and the capacitance of the capacitor to find the charge and the potential difference across each capacitor.

Formulae:

The equivalent capacitance of a series connection of capacitors,

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

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

(a) Calculation of charge q1

Since the capacitors are arranged in series, the charge on each capacitor is the same.

Thus, charge on each capacitor is given using equation (i) and (ii) as follows:

$$\begin{gathered} q=q_1 \\ =q_2 \\ =C_{eq}V \\ =\frac{C_1{C}_{2}V}{C_1+C_2} \\ =\frac{\left(2.00\mathrm{\mu F}\right)\left(8.00\mathrm{\mu F}\right)\left(300\mathrm{V}\right)}{2.00\mathrm{\mu F}+8.00\mathrm{\mu F}} \\ =4.80\times {10}^{-4}C \end{gathered}$$

Hence, the value of the charge is $4.80\times {10}^{-4}C$.

(b) Calculation of potential difference, V1

Using the similar equation (iii), the value of potential difference across capacitor 1 is given as:

$$\begin{gathered} V_1=\frac{q}{C_1} \\ =\frac{4.80\times {10}^{-4}\mathrm{C}}{2.00\mathrm{\mu F}} \\ =240\mathrm{V} \end{gathered}$$

Hence, the value of potential is 240 V.

(c) Calculation of charge, q2

From part (a) calculations, we found that the charge is same as the charge .

Hence, the value of the charge is $4.80\times {10}^{-4}C$.

(d) Calculation of potential difference, V2

The total potential difference V is the sum of the potential differences across both the capacitors. Thus, the value of the potential difference can be given as:

$$\begin{gathered} V_2=V-V_1 \\ =300V-240V \\ =60.0V \end{gathered}$$

Hence, the value of the potential is 60.0 V.

(e) Calculation of charge q1

The charged capacitor are disconnected from each other and from the battery and then the capacitors are reconnected with the plates of same signs wired together and the battery is not used, then in this case, we can treat the capacitors’ combination as parallel combination. In parallel combination, the potential difference across each capacitor is the same but the charge is different.

Let the new potential be $V\text{'}$and the new charge on $C_1$and $C_2$is ${\mathrm{q}}_1^{\text{'}}\mathrm{and}{\mathrm{q}}_2^{\text{'}}$. Thus, using equation (ii) can be related as:

$\frac{q_1^{\text{'}}}{C_1}=\frac{q_2^{\text{'}}}{C_2}$ …(iii)

And using the charge conservation law,

$q_1^{\text{'}}+q_2^{\text{'}}=q_1+q_2$

But in the first case,$q_1(=q_2)=q$

That gives the above equation as:

$q_1^{\text{'}}+q_2^{\text{'}}=2q$ …(iv)

Now, using the equation (iii) and (iv) we can find the chargeas follows:

$$\begin{gathered} {\mathrm{q}}_1^{\text{'}}+\frac{{\mathrm{q}}_1^{\text{'}}{\mathrm{C}}_2}{{\mathrm{C}}_1}=2\mathrm{q} \\ \frac{{\mathrm{q}}_1^{\text{'}}\left({\mathrm{C}}_1+{\mathrm{C}}_2\right)}{{\mathrm{C}}_1}=2\mathrm{q} \\ {\mathrm{q}}_1^{\text{'}}=\frac{2{\mathrm{qC}}_1}{{\mathrm{C}}_1+{\mathrm{C}}_2} \\ =\frac{2\left(4.80\times {10}^{-4}\mathrm{C}\right)\left(2.00\mathrm{\mu F}\right)}{2.00\mathrm{\mu F}+8.00\mathrm{\mu F}} \\ =1.92\times {10}^{-4}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $1.92\times {10}^{-4}C$.

(f) Calculation of potential difference, V1

Using the above charge value, the potential difference across capacitor 1 can be given using equation (ii) as:

$$\begin{gathered} V_1^{\text{'}}=\frac{q_1^{\text{'}}}{C_1} \\ =\frac{1.92\times {10}^{-4}C}{2.00\mathrm{\mu F}} \end{gathered}$$

Hence, the value of the potential is 96.0 V.

(g) Calculation of charge, q2

Using equation (iv) and charge of calculation of (a), we can get the charge value of the capacitor 2 as:

$$\begin{gathered} {\mathrm{q}}_2^{\text{'}}=2\mathrm{q}-{\mathrm{q}}_1^{\text{'}} \\ =2\left(4.80\times {10}^{-4}\mathrm{C}\right)-1.92\times {10}^{-4}\mathrm{C} \\ =7.68\times {10}^{-4}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $7.68\times {10}^{-4}C$.

(h) Calculation of potential difference, V2

The potential is same as the potential across capacitor 1 because both capacitors are parallel now.

Hence, from part (f), we can get the value of potential difference as 96.0 V.

(i) Calculation of charge q1

Suppose, instead, the capacitors charged in part (a) are reconnected with plates of opposite signs wired together then in this case the capacitors will simply discharge themselves.

Hence, the charge value across capacitor 1 is 0C.

(j) Calculation of potential difference, V1

Since the charge is zero, the potential is also found to be 0V from equation (i).

(k) Calculation of charge, q2

From argument in part (i), we can get the charge value across capacitor 2 as 0C.

(l) Calculation of potential difference, V2

Since the charge is zero, the potential is also found to be 0V from equation (i).