Two identical capacitors $1$ and $2$ are connected in series. The capacitor $2$ contains a dielectric slab of constant $K$ as shown. They are connected to a battery of emf $V_0\ volts$ . The dielectric slab is then removed. Let $Q_1$ and $Q_2$ be the charge stored in the capacitors before removing the slab and $Q'_1$ , and $Q'_2$ be the values after removing the slab. Then
Diffcult
Download our app for free and get started
The capacitor $1$ has capacitance $C$ and as dielectric contains in capacitor $2$ so its capacitance becomes $k C$
The net capacitance $C_{e q}=\frac{C \cdot k C}{C+k C}=\frac{C k}{1+k}$
and $Q_{e q}=C_{e q} E$
When dielectric is reemoved from capacitor $2,$ its capacitance becomes $C$.
now net capacitance $C_{e q}^{\prime}=\frac{C . C}{C+C}=\frac{C}{2}$
and $Q_{e q}^{\prime}=C_{e q}^{\prime} E=\frac{C E}{2}$
When the capacitors are connected in series, they each have the same charge as the net capacitance. Thus, $Q_{e q}=Q_{1}=Q_{2}$ and $Q_{e q}^{\prime}=Q_{1}^{\prime}=Q_{2}^{\prime}$
$\therefore \frac{Q_{1}^{\prime}}{Q_{1}}=\frac{Q_{2}^{\prime}}{Q_{2}}=\frac{2 k}{k+1}$
Download our appand get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
Similar Questions
- 1A parallel plate capacitor has a uniform electric field $E$ in the space between the plates. If the distance between the plates is $d$ and area of each plate is $A,$ the energy stored in the capacitor isView Solution
- 2A capacitor of capacitance $C$ is charged to a potential difference $V$ from a cell and then disconnected from it. A charge $+Q$ is now given to its positive plate. The potential difference across the capacitor is now $Q$View Solution
if $V < CV$
- 3The capacity of a parallel plate condenser is $10\,\mu F$ without dielectric. Dielectric of constant $2$ is used to fill half the distance between the plates, the new capacitance in $\mu F$ isView Solution
- 4In the circuit shown below $C_1=10 \,\mu F , C_2=C_3=20 \,\mu F$, and $C_4=40 \,\mu F$. If the charge on $C_1$ is $20 \,\mu C$ then potential difference between $X$ and $Y$ is ......... $V$View Solution
- 5A parallel plate capacitor whose capacitance $C$ is $14\, pF$ is charged by a battery to a potential difference $V =12\, V$ between its plates. The charging battery is now disconnected and a porcelin plate with $k =7$ is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of $..........pJ$. (Assume no friction)View Solution
- 6A $12\,pF$ capacitor is connected to a $50\,V$ battery. How much electrostatic energy is stored in the capacitorView Solution
- 7A parallel plate capacitor with plate area $A$ and plate separation $d =2 \,m$ has a capacitance of $4 \,\mu F$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K =3$ (as shown in figure) will be .........$ \mu \,F$View Solution
- 8View SolutionConsider the points lying on a straight line joining two fixed opposite charges. Between the charges there is
- 9View SolutionIn the figure a capacitor is filled with dielectrics. The resultant capacitance is
- 10A series combination of $n_1$ capacitors, each of value $C_1$ is charged by a source of potential difference $4\, V.$ When another parallel combination of $n_2$ capacitors, each of value $C_2,$ is charged by a source of potential difference $V$, it has the same (total) energy stored in it, as the first combination has. The value of $C_2,$ in terms of $C_1$ is thenView Solution