Four identical plates $1, 2, 3$ and $4$ are placed parallel to each other at equal distance as shown in the figure. Plates $1$ and $4$ are joined together and the space between $2$ and $3$ is filled with a dielectric of dielectric constant $k$ $=$ $2$. The capacitance of the system between $1$ and $3$ $\&$ $2$ and $4$ are $C_1$ and $C_2$ respectively. The ratio $\frac{{{C_1}}}{{{C_2}}}$ is
Diffcult
Download our app for free and get started
The capacitance of parallel plate capacitor is $C=\frac{A \epsilon_{0}}{d}$
The capacitance of parallel plate capacitor with dielectric is $C^{\prime}=\frac{A k \epsilon_{0}}{d}=k C=2 C$
Here $1$ and $2$ plate and $3$ and $4$ plate will make the capacitor of same capacitance $C$ and $2$ and $3$ will make the capacitance of $C^{\prime}$
Here the capacitance in between $1$ and $3$ is $C_{1}=\frac{C C^{\prime}}{C+C^{\prime}}=\frac{C .2 C}{C+2 C}=2 C / 3$
and the capacitance in between $2$ and $4$ is $C_{2}=\frac{C C^{\prime}}{C+C^{\prime}}=\frac{C .2 C}{C+2 C}=2 C / 3$
$\therefore \frac{C_{1}}{C_{2}}=1$
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 having a separation between the plates $d$ , plate area $A$ and material with dielectric constant $K$ has capacitance $C_0$. Now one-third of the material is replaced by another material with dielectric constant $2K$, so that effectively there are two capacitors one with area $\frac{1}{3}\,A$ , dielectric constant $2K$ and another with area $\frac{2}{3}\,A$ and dielectric constant $K$. If the capacitance of this new capacitor is $C$ then $\frac{C}{{{C_0}}}$ isView Solution
- 2A particle $A$ has charge $+q$ and particle $B$ has charge $+4 q$ with each of them having the same mass $m$. When allowed to fall from rest through the same electric potential difference, the ratio of their speeds $\frac{V_A}{V_B}$ will becomeView Solution
- 3Two particles each of mass $m$ and charge $q$ are separated by distance $r_1$ and the system is left free to move at $t = 0$. At time $t$ both the particles are found to be separated by distance $r_2$. The speed of each particle isView Solution
- 4View SolutionAssertion: Electron move away from a region of higher potential to a region of lower potential.
Reason: An electron has a negative charge.
- 5A capacitor of capacitance $C_1 = 1\ \mu F$ can with stand maximum voltage $V_1= 6\ kV$ (kilo-volt) and another capacitor of capacitance $ C_2 = 3\ \mu F$ can withstand maximum voltage $V_2 = 4\ kV$. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of......$kV$View Solution
- 6A capacitor with capacitance $5\,\mu F$ is charged to $5\,\mu C.$ If the plates are pulled apart to reduce the capacitance to $2\,\mu F,$ how much work is done?View Solution
- 7Two batteries and two condensers are connected as shown in the figure. The charge on $2\,\mu F$ capacitor is.....$\mu F$View Solution
- 8A charge $( - q)$ and another charge $( + Q)$ are kept at two points $A$ and $B$ respectively. Keeping the charge $( + Q)$ fixed at $B$, the charge $( - q)$ at $A$ is moved to another point $C$ such that $ABC$ forms an equilateral triangle of side $l$. The net work done in moving the charge $( - q)$ isView Solution
- 9Two capacitors of capacitances $C$ and $2\, C$ are charged to potential differences $V$ and $2\, V$, respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is$.....CV^2$View Solution
- 10In the circuit shown in figure, four capacitors are connected to a battery. The equivalent capacitance of the circuit is......$ \mu F$View Solution
