Combination of Capacitors

A network of four capacitors each of $15 \mu F$capacitance is connected to a $500 V$ supply as shown in the figure. Determine $\left(a\right)$ equivalent capacitance of the network and $\left(b\right)$ charge on $C_{4 }$capacitor.

Two capacitors of capacitance $6 \mathrm{\mu F}$ and $12 \mathrm{\mu F}$ are connected in series with a battery. The voltage across the $6 \mathrm{\mu F}$ capacitor is $2 \mathrm{V}$. The total battery voltage is

Find the equivalent capacitance between two points $A$ & $B$, for given figure (electric circuit) [Capacitance of each capacitor is $\mathrm{C}=3 \mu \mathrm{F}$]

$2 \mathrm{\mu F}$ capacitor is connected with $50 \mathrm{V}$ supply & $3 \mathrm{\mu F}$ capacitor is connected with $100 \mathrm{V}$ supply. Now after removing battery if two plates of same type of charges are placed to form new capacitor then potential difference is _____ $\mathrm{V}$.

Four plates of equal area$A$ are separated by equal distances $d$ and are arranged as shown in the figure. The equivalent capacity is

Three equal capacitors, each with capacitance $C$ are connected as shown in figure. Then the equivalent capacitance between point $A$ and $B$ is

Figure shows a network of capacitors where the number indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be $1 \mu F$ is :

The parallel combination of two air filled parallel plate capacitors of capacitance $C$ and $nC$ is connected to a battery of voltage, $V.$ When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant $K$ is placed the two plates of the first capacitor. The new potential difference of the combined system is:

A capacitance of $\mathrm{2 }\mathrm{\mu F}$ is required in an electrical circuit across a potential difference of $1\mathrm{.}0 \mathrm{kV}$. A large number of $1 \mu \mathrm{F}$ capacitors are available which can withstand a potential difference of not more than $300 \mathrm{V}$. The minimum number of capacitors required to achieve this is:

The equivalent capacitance of the combination of the capacitors is:

The charge on $3 \mathrm{\mu }\mathrm{F}$ capacitor shown in the figure is

The parallel combination of two air filled parallel plate capacitors of capacitance $C$ and $nC$ is connected to a battery of voltage, $V.$ When the capacitors are fully charged, the battery is removed and after that a dielectric material of dielectric constant $K$ is placed the two plates of the first capacitor. The new potential difference of the combined system is:

A parallel plate capacitor with plate area $\mathrm{A}$ and separation between the plates $\mathrm{d}$ is filled with different dielectrics as shown in the figure. If the equivalent capacitance between $\mathrm{a}\mathrm{ }\&\mathrm{ }\mathrm{b}$ is $\frac{m\mathrm{A}{\epsilon }_0}{n\mathrm{d}}$ then find the value of $\left(m+n\right)$, where $m$ and $n$ are least positive integers.

Five plates of area 'A' are placed very closed to each other as shown in the figure. Distance between two consecutive plates is 'd'.

Equivalent capacitance of system between the terminals $a,c$ is $\frac{p}{4}\frac{{\epsilon }_{0}A}{d}$. The value of $p$ is:

In the following circuit all capacitors have same capacitance and all are neutral except capacitor $1,$ which has charge on it. When switch is closed then in steady state it has charge of $4\mathrm{\mu C}$ then if the initial charge on it was $\frac{x}{5} \mathrm{\mu C}$, what is the value of $x$?

Two identical parallel plate capacitors are connected in parallel and charged to a voltage $V_0=12 \mathrm{V}.$ When you disconnect them from the voltage source and the distance between the plates of one of the capacitors is reduced to one third of the original distance, what will be the new voltage on the capacitors (in volts) ?

Three condensers are connected as shown in series. The insulated plate of $C_1$ is at $45 V$ and right plate of $C_3$ is earthed. If the potential difference between the plates of $C_2$ is $2n$ Volt, find value of $n$.

After the switch is closed, find potential difference between the plates $Q$ and $S.$

There are six parallel plates of equal area $A$ are arranged as shown in figure $\left(\mathrm{d}<<\mathrm{A}\right).$ The equivalent capacitance between points $2$ and $5,$ is $\alpha \frac{{\epsilon }_0 A}{\mathit{ }d}.$ Then find the value of $\alpha$.

Three capacitors $2 \mathrm{\mu F}, 3 \mathrm{\mu F}$ and $6 \mathrm{\mu }\mathrm{F}$ are joined with each other. What is the minimum effective capacitance (in $\mathrm{\mu F}$)?