Question

A capacitor of capacitance ${\mathbf{C}}_{\mathbf{1}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu F}$is connected in series with a capacitor of capacitance${\mathbf{C}}_2\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\mu F}$, and a potential difference of 200 Vis applied across the pair. (a) Calculate the equivalent capacitance. What are (b) charge${\mathit{q}}_{\mathbf{1}}$and (c) potential difference${\mathit{V}}_{\mathbf{1}}$ on capacitor 1 and (d)${\mathit{q}}_{\mathbf{2}}$and (e)${\mathit{V}}_{\mathbf{2}}$on capacitor 2?

Short answer

1. Equivalent capacitance is $2.40\mu F$.
2. Charge $q_1$is$4.80\times {10}^{-4}C$
3. Potential difference$V_1$ on capacitor 1 is 80 V.
4. Charge $q_2$is$4.80\times {10}^{-4}C$
5. Potential difference$V_2$ on capacitor 2 is 120 V.

Step-by-step solution

The given data

1. Capacitance of the capacitor 1,$C_1=6.00\mu F$
2. Capacitance of capacitor 2,$C_2=4.00\mu F$

c. Potential difference across the pair, V = 200 V

Understanding the concept of the equivalent capacitance

We first find the equivalent capacitance of the series combination and then using the relation of charge and capacitance we find the charge on each capacitor and the potential difference across it.

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 the equivalent capacitance

Since two capacitors are in series their equivalent capacitance is given using equation (i) by,

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2} \\ {C}_{eq}=\frac{C_1{C}_2}{C_1+C_2} \\ =\frac{6\mu F\times 4\mu F}{6\mu F+4\mu F} \\ =2.40\mu F \end{gathered}$$

Hence, the value of the capacitance is $2.40\mu F$.

(b) Calculation of the charge, q1

The value of the charge using the potential difference across the battery can be given using equation (ii) as:

$\begin{array}{ll}{q}_1=2.40\mu F\times 200& V\\ =480\mu C& \\ =4.80\times {10}^{-4}C& \end{array}$

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

(c) Calculation of the potential difference, V1

The value of the potential acrossis given using the given data in equation (ii) as follows:

$\begin{array}{l}{V}_1=\frac{4.80\times {10}^{-4}C}{6.00\mu F^F}\\ =80.0V\end{array}$

Hence, the value of the potential is 80.0 V.

(d) Calculation of the charge, q2

Since the two capacitors are in series the charge across each capacitor is same. Hence the charge on $C_1$and $C_2$is same, which is $4.80\times {10}^{-4}C$.

(e) Calculation of the potential difference, V2

Total potential difference V is the sum of the potential difference across $C_1$and$C_2$.

Thus, the value of the potential difference$V_2$ is given using the data in equation (ii) as follows:

$\begin{array}{l}{V}_2=V-V_1\\ =200V-80.0V\\ =120V\end{array}$

Hence, the value of the potential difference is 120 V.